A. Sakurako's Exam
模拟，两个 2 能够自己变成 0，因此将 b 对 2 取余；多余的 2 还能跟两个 1 组合在一起变成 0；最后判断剩余的 1 的数量是奇数还是偶数。
cpp
#include <bits/stdc++.h>
using namespace std;
inline int read() {
int x = 0, f = 1; char c = getchar();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int main() {
int t = read();
while (t--) {
int a = read(), b = read();
b %= 2;
while (a >= 2 && b) {
a -= 2;
b--;
}
if (b || (a % 2 == 1)) puts("NO");
else puts("YES");
}
return 0;
}B. Square or Not
首先，n 必须是一个完全平方数。为了实现起来较为简单，可以先构造一个期望的矩阵，再将矩阵转换成对应的字符串，最后判断该字符串是否与输入的字符串一致。
cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
string s, str;
inline int read() {
int x = 0, f = 1; char c = getchar();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
void solve() {
int n = read();
cin >> s;
str = "";
int num = sqrt(n);
if (num * num < n) { puts("No"); return; }
char a[num + 5][num + 5];
for (int i = 1; i <= num; i++)
for (int j = 1; j <= num; j++) a[i][j] = '0';
for (int i = 1; i <= num; i++) a[1][i] = a[num][i] = a[i][1] = a[i][num] = '1';
for (int i = 1; i <= num; i++)
for (int j = 1; j <= num; j++) str += a[i][j];
if (s == str) puts("Yes");
else puts("No");
}
int main() {
int t = read();
while (t--) {
solve();
}
return 0;
}C. Longest Good Array
题目要求的数列是数列本身单调递增，并且数列的差单调递增。要使数列最长，就要使数列的差尽可能的小，那么构造出来数列的差一定是 1，2，...，n。数列的第一位是 ，最后一位是
。所以一定要满足
，即
。
我们只需要二分 n，n + 1 即为最大长度。
cpp
#include <bits/stdc++.h>
using namespace std;
#define int long long
inline int read() {
int x = 0, f = 1; char c = getchar();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int solve() {
int x = read(), y = read();
int l = 1, r = y - x, ans = 0;
while (l <= r) {
int mid = l + r >> 1;
if (mid * (mid + 1) / 2 <= y - x) ans = mid, l = mid + 1;
else r = mid - 1;
}
return ans;
}
signed main() {
int t = read();
while (t--) {
printf("%lld\n", solve() + 1);
}
return 0;
}D. Sakurako's Hobby
可以将该问题抽象成一张图，利用图论的知识完成。
- 如果 i 不等于 p[i]，那么我们建一条由 i 指向 p[i] 的有向边（可能会有环）。
- 进行拓扑排序，由于图中可能会有环，所以无法得到全部节点的答案。
- 我们很容易发现，同一个环中的每一个节点答案都是一样的，利用 dfs 将同一个环中的所有节点存入一个 vector 中并且记录该环的答案，再将答案填到对应节点。
cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n, p[N], f[N], head[N], in[N], vis[N], num = 0, sum;
string s;
struct edge { int to, nxt; } e[N];
vector<int> circle;
inline int read() {
int x = 0, f = 1; char c = getchar();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
void addEdge(int u, int v) { e[++num] = (edge){v, head[u]}, head[u] = num, in[v]++; }
void dfs(int u) {
circle.push_back(u); vis[u] = 1;
if (s[u] == '0') sum++;
for (int i = head[u], v; i; i = e[i].nxt) {
v = e[i].to;
if (!vis[v]) dfs(v);
}
}
int main() {
int t = read();
while (t--) {
n = read();
num = 0;
for (int i = 1; i <= n; i++) p[i] = read(), head[i] = in[i] = vis[i] = f[i] = 0;
for (int i = 1; i <= n; i++) {
if (i != p[i]) addEdge(i, p[i]);
}
cin >> s;
s = " " + s;
queue<int> q;
for (int i = 1; i <= n; i++) {
if (!in[i]) q.push(i), vis[i] = 1;
}
while (!q.empty()) {
int u = q.front(); q.pop();
if (s[u] == '0') f[u]++;
for (int i = head[u], v; i; i = e[i].nxt) {
v = e[i].to; in[v]--;
f[v] += f[u];
if (!in[v] && !vis[v]) q.push(v), vis[v] = 1;
}
}
for (int i = 1; i <= n; i++) {
if (!vis[i]) {
circle.clear();
sum = 0;
dfs(i);
for (int j = 0; j < circle.size(); j++) f[circle[j]] = sum;
}
}
for (int i = 1; i <= n; i++) {
printf("%d", f[i]);
if (i < n) putchar(' ');
else putchar('\n');
}
}
return 0;
}F. Sakurako's Box
暴力求出所有乘积，最后利用乘法逆元得出结果。
cpp
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 10, mod = 1e9 + 7;
int n, pre[N], a[N], P, Q;
inline int read() {
int x = 0, f = 1; char c = getchar();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int qpow(int x, int k) {
int res = 1LL;
while (k) {
if (k & 1) res = res * x % mod;
x = x * x % mod;
k >>= 1;
}
return res % mod;
}
signed main() {
int t = read();
while (t--) {
n = read();
for (int i = 1; i <= n; i++) a[i] = read(), pre[i] = (pre[i - 1] + a[i]) % mod;
Q = (n * (n - 1) / 2) % mod, P = 0;
for (int i = 1; i <= n; i++) P = (P + a[i] * pre[i - 1] % mod) % mod;
printf("%lld\n", qpow(Q, mod - 2) * P % mod);
}
return 0;
}