A - The bottom of the ninth
Problem Statement
Team Takahashi and Team Aoki are playing a baseball game, with Team Takahashi batting first.
Currently, the game has finished through the top of the ninth inning, and the bottom of the ninth is about to begin.
Team Takahashi scored A i A_i Ai runs in the top of the i i i-th inning ( 1 ≤ i ≤ 9 ) (1\leq i\leq 9) (1≤i≤9), and Team Aoki scored B j B_j Bj runs in the bottom of the j j j-th inning ( 1 ≤ j ≤ 8 ) (1\leq j\leq 8) (1≤j≤8).
At the end of the top of the ninth, Team Takahashi's score is not less than Team Aoki's score.
Determine the minimum number of runs Team Aoki needs to score in the bottom of the ninth to win the game.
Here, if the game is tied at the end of the bottom of the ninth, it results in a draw. Therefore, for Team Aoki to win, they must score strictly more runs than Team Takahashi by the end of the bottom of the ninth.
Team Takahashi's score at any point is the total runs scored in the tops of the innings up to that point, and Team Aoki's score is the total runs scored in the bottoms of the innings.
Constraints
0 ≤ A i , B j ≤ 99 0\leq A_i, B_j\leq 99 0≤Ai,Bj≤99
A 1 + A 2 + A 3 + A 4 + A 5 + A 6 + A 7 + A 8 + A 9 ≥ B 1 + B 2 + B 3 + B 4 + B 5 + B 6 + B 7 + B 8 A_1 + A_2 + A_3 + A_4 + A_5 + A_6 + A_7 + A_8 + A_9 \geq B_1 + B_2 + B_3 + B_4 + B_5 + B_6 + B_7 + B_8 A1+A2+A3+A4+A5+A6+A7+A8+A9≥B1+B2+B3+B4+B5+B6+B7+B8
All input values are integers.
Input
The input is given from Standard Input in the following format:
A 1 A_1 A1 A 2 A_2 A2 A 3 A_3 A3 A 4 A_4 A4 A 5 A_5 A5 A 6 A_6 A6 A 7 A_7 A7 A 8 A_8 A8 A 9 A_9 A9
B 1 B_1 B1 B 2 B_2 B2 B 3 B_3 B3 B 4 B_4 B4 B 5 B_5 B5 B 6 B_6 B6 B 7 B_7 B7 B 8 B_8 B8
Output
Print the minimum number of runs Team Aoki needs to score in the bottom of the ninth inning to win.
Sample Input 1
0 1 0 1 2 2 0 0 1
1 1 0 0 0 0 1 0
Sample Output 1
5
At the end of the top of the ninth inning, Team Takahashi has scored seven runs, and Team Aoki has scored three runs.
Therefore, if Team Aoki scores five runs in the bottom of the ninth, the scores will be 7 − 8 7-8 7−8, allowing them to win.
Note that scoring four runs would result in a draw and not a victory.
Sample Input 2
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
Sample Output 2
1
Solution
具体见文末视频。
Code
cpp
#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
int s1 = 0, s2 = 0, x;
for (int i = 1; i <= 9; i ++)
cin >> x, s1 += x;
for (int i = 1; i <= 8; i ++)
cin >> x, s2 += x;
if (s1 < s2) {
cout << 0 << endl;
} else {
cout << s1 - s2 + 1 << endl;
}
return 0;
}B - Spot the Difference
Problem Statement
You are given two grids, each with N N N rows and N N N columns, referred to as grid A A A and grid B B B.
Each cell in the grids contains a lowercase English letter.
The character at the i i i-th row and j j j-th column of grid A A A is A i , j A_{i, j} Ai,j.
The character at the i i i-th row and j j j-th column of grid B B B is B i , j B_{i, j} Bi,j.
The two grids differ in exactly one cell. That is, there exists exactly one pair ( i , j ) (i, j) (i,j) of positive integers not greater than N N N such that A i , j ≠ B i , j A_{i, j} \neq B_{i, j} Ai,j=Bi,j. Find this ( i , j ) (i, j) (i,j).
Constraints
1 ≤ N ≤ 100 1 \leq N \leq 100 1≤N≤100
A i , j A_{i, j} Ai,j and B i , j B_{i, j} Bi,j are all lowercase English letters.
There exists exactly one pair ( i , j ) (i, j) (i,j) such that A i , j ≠ B i , j A_{i, j} \neq B_{i, j} Ai,j=Bi,j.
Input
The input is given from Standard Input in the following format:
N N N
A 1 , 1 A 1 , 2 ... A 1 , N A_{1,1}A_{1,2}\dots A_{1,N} A1,1A1,2...A1,N
A 2 , 1 A 2 , 2 ... A 2 , N A_{2,1}A_{2,2}\dots A_{2,N} A2,1A2,2...A2,N
⋮ \vdots ⋮
A N , 1 A N , 2 ... A N , N A_{N,1}A_{N,2}\dots A_{N,N} AN,1AN,2...AN,N
B 1 , 1 B 1 , 2 ... B 1 , N B_{1,1}B_{1,2}\dots B_{1,N} B1,1B1,2...B1,N
B 2 , 1 B 2 , 2 ... B 2 , N B_{2,1}B_{2,2}\dots B_{2,N} B2,1B2,2...B2,N
⋮ \vdots ⋮
B N , 1 B N , 2 ... B N , N B_{N,1}B_{N,2}\dots B_{N,N} BN,1BN,2...BN,N
Output
Let ( i , j ) (i, j) (i,j) be the pair of positive integers not greater than N N N such that A i , j ≠ B i , j A_{i, j} \neq B_{i, j} Ai,j=Bi,j. Print ( i , j ) (i, j) (i,j) in the following format:
i i i j j j
Sample Input 1
3
abc
def
ghi
abc
bef
ghi
Sample Output 1
2 1
From A 2 , 1 = A_{2, 1} = A2,1= d and B 2 , 1 = B_{2, 1} = B2,1= b, we have A 2 , 1 ≠ B 2 , 1 A_{2, 1} \neq B_{2, 1} A2,1=B2,1, so ( i , j ) = ( 2 , 1 ) (i, j) = (2, 1) (i,j)=(2,1) satisfies the condition in the problem statement.
Sample Input 2
1
f
q
Sample Output 2
1 1
Sample Input 3
10
eixfumagit
vtophbepfe
pxbfgsqcug
ugpugtsxzq
bvfhxyehfk
uqyfwtmglr
jaitenfqiq
acwvufpfvv
jhaddglpva
aacxsyqvoj
eixfumagit
vtophbepfe
pxbfgsqcug
ugpugtsxzq
bvfhxyehok
uqyfwtmglr
jaitenfqiq
acwvufpfvv
jhaddglpva
aacxsyqvoj
Sample Output 3
5 9
Solution
具体见文末视频。
Code
cpp
#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
const int N = 1e2 + 10;
int n;
char a[N][N], b[N][N];
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
cin >> n;
for (int i = 1; i <= n; i ++)
for (int j = 1; j <= n; j ++)
cin >> a[i][j];
for (int i = 1; i <= n; i ++)
for (int j = 1; j <= n; j ++)
cin >> b[i][j];
for (int i = 1; i <= n; i ++)
for (int j = 1; j <= n; j ++)
if (a[i][j] != b[i][j]) {
cout << i << " " << j << endl;
return 0;
}
return 0;
}C - Merge the balls
Problem Statement
You have an empty sequence and N N N balls. The size of the i i i-th ball ( 1 ≤ i ≤ N ) (1 \leq i \leq N) (1≤i≤N) is 2 A i 2^{A_i} 2Ai.
You will perform N N N operations.
In the i i i-th operation, you add the i i i-th ball to the right end of the sequence, and repeat the following steps:
- If the sequence has one or fewer balls, end the operation. If the rightmost ball and the second rightmost ball in the sequence have different sizes, end the operation. If the rightmost ball and the second rightmost ball in the sequence have the same size, remove these two balls and add a new ball to the right end of the sequence with a size equal to the sum of the sizes of the two removed balls. Then, go back to step 1 and repeat the process.
Determine the number of balls remaining in the sequence after the N operations. #### Constraints
1 ≤ N ≤ 2 × 1 0 5 1 \leq N \leq 2 \times 10^5 1≤N≤2×105
0 ≤ A i ≤ 1 0 9 0 \leq A_i \leq 10^9 0≤Ai≤109
All input values are integers.
Input
The input is given from Standard Input in the following format:
N N N
A 1 A_1 A1 A 2 A_2 A2 ... \ldots ... A N A_N AN
Output
Print the number of balls in the sequence after the N N N operations.
Sample Input 1
7
2 1 1 3 5 3 3
Sample Output 1
3
The operations proceed as follows:
After the first operation, the sequence has one ball, of size 2 2 2^2 22.
After the second operation, the sequence has two balls, of sizes 2 2 2^2 22 and 2 1 2^1 21 in order.
After the third operation, the sequence has one ball, of size 2 3 2^3 23. This is obtained as follows:
When the third ball is added during the third operation, the sequence has balls of sizes 2 2 , 2 1 , 2 1 2^2, 2^1, 2^1 22,21,21 in order.
The first and second balls from the right have the same size, so these balls are removed, and a ball of size 2 1 + 2 1 = 2 2 2^1 + 2^1 = 2^2 21+21=22 is added. Now, the sequence has balls of sizes 2 2 , 2 2 2^2, 2^2 22,22.
Again, the first and second balls from the right have the same size, so these balls are removed, and a ball of size 2 2 + 2 2 = 2 3 2^2 + 2^2 = 2^3 22+22=23 is added, leaving the sequence with a ball of size 2 3 2^3 23.
After the fourth operation, the sequence has one ball, of size 2 4 2^4 24.
After the fifth operation, the sequence has two balls, of sizes 2 4 2^4 24 and 2 5 2^5 25 in order.
After the sixth operation, the sequence has three balls, of sizes 2 4 2^4 24, 2 5 2^5 25, 2 3 2^3 23 in order.
After the seventh operation, the sequence has three balls, of sizes 2 4 2^4 24, 2 5 2^5 25, 2 4 2^4 24 in order.
Therefore, you should print 3 3 3, the final number of balls in the sequence.
Sample Input 2
5
0 0 0 1 2
Sample Output 2
4
The operations proceed as follows:
After the first operation, the sequence has one ball, of size 2 0 2^0 20.
After the second operation, the sequence has one ball, of size 2 1 2^1 21.
After the third operation, the sequence has two balls, of sizes 2 1 2^1 21 and 2 0 2^0 20 in order.
After the fourth operation, the sequence has three balls, of sizes 2 1 2^1 21, 2 0 2^0 20, 2 1 2^1 21 in order.
After the fifth operation, the sequence has four balls, of sizes 2 1 2^1 21, 2 0 2^0 20, 2 1 2^1 21, 2 2 2^2 22 in order.
Therefore, you should print 4 4 4, the final number of balls in the sequence.
Solution
具体见文末视频。
Code
cpp
#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
int n;
cin >> n;
std::vector<int> b;
int x;
for (int i = 1; i <= n; i ++) {
cin >> x;
b.emplace_back(x);
while (b.size() > 1 && b.back() == b[b.size() - 2]) {
int tmp = b.back();
b.pop_back(), b.pop_back();
b.emplace_back(tmp + 1);
}
}
cout << b.size() << endl;
return 0;
}D - Grid and Magnet
Problem Statement
There is a grid of H H H rows and W W W columns. Some cells (possibly zero) contain magnets.
The state of the grid is represented by H H H strings S 1 , S 2 , ... , S H S_1, S_2, \ldots, S_H S1,S2,...,SH of length W W W. If the j j j-th character of S i S_i Si is #, it indicates that there is a magnet in the cell at the i i i-th row from the top and j j j-th column from the left; if it is ., it indicates that the cell is empty.
Takahashi, wearing an iron armor, can move in the grid as follows:
If any of the cells vertically or horizontally adjacent to the current cell contains a magnet, he cannot move at all.
Otherwise, he can move to any one of the vertically or horizontally adjacent cells.
However, he cannot exit the grid.
For each cell without a magnet, define its degree of freedom as the number of cells he can reach by repeatedly moving from that cell. Find the maximum degree of freedom among all cells without magnets in the grid.
Here, in the definition of degree of freedom, "cells he can reach by repeatedly moving" mean cells that can be reached from the initial cell by some sequence of moves (possibly zero moves). It is not necessary that there is a sequence of moves that visits all such reachable cells starting from the initial cell. Specifically, each cell itself (without a magnet) is always included in the cells reachable from that cell.
Constraints
1 ≤ H , W ≤ 1000 1 \leq H, W \leq 1000 1≤H,W≤1000
H H H and W W W are integers.
S i S_i Si is a string of length W W W consisting of . and #.
There is at least one cell without a magnet.
Input
The input is given from Standard Input in the following format:
H H H W W W
S 1 S_1 S1
S 2 S_2 S2
⋮ \vdots ⋮
S H S_H SH
Output
Print the maximum degree of freedom among all cells without magnets.
Sample Input 1
3 5
.#...
.....
.#..#
Sample Output 1
9
Let ( i , j ) (i,j) (i,j) denote the cell at the i i i-th row from the top and j j j-th column from the left. If Takahashi starts at ( 2 , 3 ) (2,3) (2,3), possible movements include:
( 2 , 3 ) → ( 2 , 4 ) → ( 1 , 4 ) → ( 1 , 5 ) → ( 2 , 5 ) (2,3) \to (2,4) \to (1,4) \to (1,5) \to (2,5) (2,3)→(2,4)→(1,4)→(1,5)→(2,5)
( 2 , 3 ) → ( 2 , 4 ) → ( 3 , 4 ) (2,3) \to (2,4) \to (3,4) (2,3)→(2,4)→(3,4)
( 2 , 3 ) → ( 2 , 2 ) (2,3) \to (2,2) (2,3)→(2,2)
( 2 , 3 ) → ( 1 , 3 ) (2,3) \to (1,3) (2,3)→(1,3)
( 2 , 3 ) → ( 3 , 3 ) (2,3) \to (3,3) (2,3)→(3,3)
Thus, including the cells he passes through, he can reach at least nine cells from ( 2 , 3 ) (2,3) (2,3).
Actually, no other cells can be reached, so the degree of freedom for ( 2 , 3 ) (2,3) (2,3) is 9 9 9.
This is the maximum degree of freedom among all cells without magnets, so print 9 9 9.
Sample Input 2
3 3
..#
#..
..#
Sample Output 2
1
For any cell without a magnet, there is a magnet in at least one of the adjacent cells.
Thus, he cannot move from any of these cells, so their degrees of freedom are 1 1 1.
Therefore, print 1 1 1.
Solution
具体见文末视频。
Code
cpp
#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
const int N = 1e3 + 10;
int n, m;
char g[N][N];
int vis[N][N], dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1}, st[N][N];
PII q[N * N];
int hh, tt;
int bfs(int sx, int sy) {
hh = 0, tt = 0;
int res = 1;
queue<PII> qt;
q[0] = {sx, sy}, st[sx][sy] = 1, vis[sx][sy] = 1;
qt.emplace(sx, sy);
while (hh <= tt) {
auto tmp = q[hh ++];
bool flg = 1;
for (int i = 0; i < 4; i ++) {
int xx = tmp.fi + dx[i], yy = tmp.se + dy[i];
if (xx < 1 || yy < 1 || xx > n || yy > m) continue;
if (g[xx][yy] == '#') {
flg = 0;
break;
}
}
if (flg) {
for (int i = 0; i < 4; i ++) {
int xx = tmp.fi + dx[i], yy = tmp.se + dy[i];
if (xx < 1 || yy < 1 || xx > n || yy > m) continue;
if (!st[xx][yy]) res ++, st[xx][yy] = 1, qt.emplace(xx, yy);
if (!vis[xx][yy]) q[ ++ tt] = {xx, yy}, vis[xx][yy] = 1;
}
}
}
while (qt.size()) {
auto v = qt.front();
qt.pop();
st[v.fi][v.se] = 0;
}
return res;
}
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
cin >> n >> m;
for (int i = 1; i <= n; i ++)
for (int j = 1; j <= m; j ++)
cin >> g[i][j];
int res = 0;
for (int i = 1; i <= n; i ++)
for (int j = 1; j <= m; j ++)
if (!vis[i][j] && g[i][j] == '.') {
int val = bfs(i, j);
res = max(res, val);
// cout << i << " " << j << ":" << val << endl;
}
cout << res << endl;
return 0;
}E - Jump Distance Sum
Problem Statement
On a coordinate plane, there are N N N points P 1 , P 2 , ... , P N P_1, P_2, \ldots, P_N P1,P2,...,PN, where point P i P_i Pi has coordinates ( X i , Y i ) (X_i, Y_i) (Xi,Yi).
The distance dist ( A , B ) \text{dist}(A, B) dist(A,B) between two points A A A and B B B is defined as follows:
A rabbit is initially at point A.
A rabbit at position (x, y) can jump to (x+1, y+1), (x+1, y-1), (x-1, y+1), or (x-1, y-1) in one jump.
\\text{dist}(A, B) is defined as the minimum number of jumps required to get from point A to point B.
If it is impossible to get from point A to point B after any number of jumps, let \\text{dist}(A, B) = 0.
Calculate the sum \\displaystyle\\sum_{i=1}\^{N-1}\\displaystyle\\sum_{j=i+1}\^N \\text{dist}(P_i, P_j). #### Constraints
2 ≤ N ≤ 2 × 1 0 5 2 \leq N \leq 2 \times 10^5 2≤N≤2×105
0 ≤ X i , Y i ≤ 1 0 8 0 \leq X_i, Y_i \leq 10^8 0≤Xi,Yi≤108
For i ≠ j i \neq j i=j, ( X i , Y i ) ≠ ( X j , Y j ) (X_i, Y_i) \neq (X_j, Y_j) (Xi,Yi)=(Xj,Yj)
All input values are integers.
Input
The input is given from Standard Input in the following format:
N N N
X 1 X_1 X1 Y 1 Y_1 Y1
X 2 X_2 X2 Y 2 Y_2 Y2
⋮ \vdots ⋮
X N X_N XN Y N Y_N YN
Output
Print the value of ∑ i = 1 N − 1 ∑ j = i + 1 N dist ( P i , P j ) \displaystyle\sum_{i=1}^{N-1}\displaystyle\sum_{j=i+1}^N \text{dist}(P_i, P_j) i=1∑N−1j=i+1∑Ndist(Pi,Pj) as an integer.
Sample Input 1
3
0 0
1 3
5 6
Sample Output 1
3
P 1 P_1 P1, P 2 P_2 P2, and P 3 P_3 P3 have coordinates ( 0 , 0 ) (0,0) (0,0), ( 1 , 3 ) (1,3) (1,3), and ( 5 , 6 ) (5,6) (5,6), respectively.
The rabbit can get from P 1 P_1 P1 to P 2 P_2 P2 in three jumps via ( 0 , 0 ) → ( 1 , 1 ) → ( 0 , 2 ) → ( 1 , 3 ) (0,0) \to (1,1) \to (0,2) \to (1,3) (0,0)→(1,1)→(0,2)→(1,3), but not in two or fewer jumps,
so dist ( P 1 , P 2 ) = 3 \text{dist}(P_1, P_2) = 3 dist(P1,P2)=3.
The rabbit cannot get from P 1 P_1 P1 to P 3 P_3 P3 or from P 2 P_2 P2 to P 3 P_3 P3, so dist ( P 1 , P 3 ) = dist ( P 2 , P 3 ) = 0 \text{dist}(P_1, P_3) = \text{dist}(P_2, P_3) = 0 dist(P1,P3)=dist(P2,P3)=0.
Therefore, the answer is ∑ i = 1 2 ∑ j = i + 1 3 dist ( P i , P j ) = dist ( P 1 , P 2 ) + dist ( P 1 , P 3 ) + dist ( P 2 , P 3 ) = 3 + 0 + 0 = 3 \displaystyle\sum_{i=1}^{2}\displaystyle\sum_{j=i+1}^3\text{dist}(P_i, P_j)=\text{dist}(P_1, P_2)+\text{dist}(P_1, P_3)+\text{dist}(P_2, P_3)=3+0+0=3 i=1∑2j=i+1∑3dist(Pi,Pj)=dist(P1,P2)+dist(P1,P3)+dist(P2,P3)=3+0+0=3.
Sample Input 2
5
0 5
1 7
2 9
3 8
4 6
Sample Output 2
11
Solution
具体见文末视频。
Code
cpp
#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
const int N = 2e5 + 10;
int n;
PII p[N];
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
cin >> n;
for (int i = 1; i <= n; i ++) {
int x, y;
cin >> x >> y;
p[i].fi = x + y, p[i].se = y - x;
}
sort(p + 1, p + 1 + n);
int res = 0;
int pcnt[2] = {0}, psum[2] = {0};
for (int i = 1; i <= n; i ++) {
int tmp = p[i].fi & 1;
res += pcnt[tmp] * p[i].fi - psum[tmp];
pcnt[tmp] ++, psum[tmp] += p[i].fi;
}
sort(p + 1, p + 1 + n, [&](PII a, PII b) {
return a.se < b.se;
});
memset(pcnt, 0, sizeof pcnt);
memset(psum, 0, sizeof psum);
for (int i = 1; i <= n; i ++) {
int tmp = p[i].fi & 1;
res += pcnt[tmp] * p[i].se - psum[tmp];
pcnt[tmp] ++, psum[tmp] += p[i].se;
}
cout << res / 2 << endl;
return 0;
}F - Double Sum
Problem Statement
You are given an integer sequence A = ( A 1 , A 2 , ... , A N ) A = (A_1, A_2, \dots, A_N) A=(A1,A2,...,AN).
Calculate the following expression:
\\displaystyle \\sum_{i=1}\^N \\sum_{j=i+1}\^N \\max(A_j - A_i, 0)
The constraints guarantee that the answer is less than 2\^{63}. #### Constraints
2 ≤ N ≤ 4 × 1 0 5 2 \leq N \leq 4 \times 10^5 2≤N≤4×105
0 ≤ A i ≤ 1 0 8 0 \leq A_i \leq 10^8 0≤Ai≤108
All input values are integers.
Input
The input is given from Standard Input in the following format:
N N N
A 1 A_1 A1 A 2 A_2 A2 ... \dots ... A N A_N AN
Output
Print the value of the expression.
Sample Input 1
3
2 5 3
Sample Output 1
4
For ( i , j ) = ( 1 , 2 ) (i, j) = (1, 2) (i,j)=(1,2), we have max ( A j − A i , 0 ) = max ( 3 , 0 ) = 3 \max(A_j - A_i, 0) = \max(3, 0) = 3 max(Aj−Ai,0)=max(3,0)=3.
For ( i , j ) = ( 1 , 3 ) (i, j) = (1, 3) (i,j)=(1,3), we have max ( A j − A i , 0 ) = max ( 1 , 0 ) = 1 \max(A_j - A_i, 0) = \max(1, 0) = 1 max(Aj−Ai,0)=max(1,0)=1.
For ( i , j ) = ( 2 , 3 ) (i, j) = (2, 3) (i,j)=(2,3), we have max ( A j − A i , 0 ) = max ( − 2 , 0 ) = 0 \max(A_j - A_i, 0) = \max(-2, 0) = 0 max(Aj−Ai,0)=max(−2,0)=0.
Adding these together gives 3 + 1 + 0 = 4 3 + 1 + 0 = 4 3+1+0=4, which is the answer.
Sample Input 2
10
5 9 3 0 4 8 7 5 4 0
Sample Output 2
58
Solution
后期补一下这题目的视频
Code
cpp
#include <bits/stdc++.h>
#define fi first
#define se second
#define int long long
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
const int N = 4e5 + 10;
int n;
PII a[N];
struct fenwick
{
int tr[N];
void add(int x, int d) {
for (int i = x; i < N; i += (i & -i)) tr[i] += d;
}
int sum(int x) {
int res = 0;
for (int i = x; i; i -= (i & -i)) res += tr[i];
return res;
}
}sum, cnt;
signed main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
cin >> n;
for (int i = 1; i <= n; i ++)
cin >> a[i].fi, a[i].se = i;
sort(a + 1, a + 1 + n);
int res = 0;
for (int i = 1; i <= n; i ++) {
res += cnt.sum(a[i].se) * a[i].fi - sum.sum(a[i].se);
cnt.add(a[i].se, 1), sum.add(a[i].se, a[i].fi);
}
cout << res << endl;
return 0;
}视频题解
Atcoder Beginner Contest 351（A ~ F 题讲解）
最后祝大家早日