给出数组 a a a，如果二元组 ( i , j ) (i,j) (i,j)满足 a i + a j m o d x = = 0 & & a i − a j m o d y = = 0 a_i + a_j mod x ==0 \&\& a_i - a_j mod y == 0 ai+ajmodx==0&&ai−ajmody==0，则beauty。其中 i < j i<j i<j
根据题意不难得出，符合条件的二元组应满足
a i m o d x + a j m o d x = x a i m o d y = a j m o d y a_i \mod x + a_j \mod x = x \\ a_i \mod y = a_j \mod y aimodx+ajmodx=xaimody=ajmody
所以用 ( a i m o d x , a i m o d y ) (a_i \mod x, a_i \mod y) (aimodx,aimody)作为key，对于每个元素 a i a_i ai查找 ( x − a i m o d x , a i m o d y ) (x- a_i \mod x, a_i \mod y) (x−aimodx,aimody)的个数
c++
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <time.h>
#include <set>
#include <map>
#include <queue>
#define IOS ios::sync_with_stdio(0);cin.tie(0);
#define rep(index,start,end) for(int index = start;index < end; index ++)
#define drep(index,start,end) for(int index = start;index >= end; index --)
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b){
ll t;
while(b){
t = b;
b = a % b;
a = t;
}
return a;
}
const int maxn = 2e5+5;
typedef pair<int, int> key;
int n,x,y;
map<key, int> store;
int main() {
IOS
int t;
cin>>t;
while(t--) {
store.clear();
cin>>n>>x>>y;
int tmp;
ll sum = 0LL;
rep(i,0,n) {
cin>>tmp;
int modx = tmp % x;
int mody = tmp % y;
key now ={modx, mody};
// cal
int bmody = tmp % y;
int bmodx = (x - tmp%x) % x;
sum += store[{bmodx, bmody}];
if (store.find(now) != store.end()) {
store[now] += 1;
} else {
store[now] = 1;
}
}
cout<<sum<<endl;
}
return 0;
}E. Anna and the Valentine's Day Gift 博弈论
俩人玩游戏，一个能选一个数reverse，一个能选一个数拼接，看最后的结果能不能大于 1 0 m 10^m 10m
如果想要减少最终结果的位数，那么必须reverse之后产生前导零，例如10000，反转后变成1。那么这道题就变成了对反转后产生前导零个数的排序。注意我们的对手不傻，当我们把产生最多前导零的数字反转后，对手肯定会把产生第二多的拼接保护，防止最终结果位数减少，所以只能减去排序结果的偶数位
c++
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <time.h>
#include <set>
#include <map>
#include <queue>
#define IOS ios::sync_with_stdio(0);cin.tie(0);
#define rep(index,start,end) for(int index = start;index < end; index ++)
#define drep(index,start,end) for(int index = start;index >= end; index --)
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b){
ll t;
while(b){
t = b;
b = a % b;
a = t;
}
return a;
}
const int maxn = 2e5+5;
struct node{
int digit;
int zero;
};
int n,m;
node store[maxn];
bool cmp(const node& a, const node& b) {
return a.zero > b.zero;
}
int main() {
IOS
int t;
cin>>t;
while(t--) {
int tmp;
cin>>n>>m;
int sum = 0;
rep(i,0,n) {
cin>>tmp;
int dig = 0;
int zero = 0;
bool leading = true;
while(tmp > 0) {
dig ++;
if (leading && !(tmp % 10)) {
zero ++;
} else {
leading = false;
}
tmp /= 10;
}
// cout<<"dig :"<<dig<<" zero:"<<zero<<endl;
store[i].digit = dig;
store[i].zero = zero;
sum += dig;
}
sort(store, store+n, cmp);
// cout<<"test log:"<<store[0].zero<<endl;
for(int i=0;i<n;i+=2) {
sum -= store[i].zero;
}
if (sum < m+1) {
cout<<"Anna"<<endl;
} else {
cout<<"Sasha"<<endl;
}
}
return 0;
}https://codeforces.com/contest/1931/problem/F 拓扑排序
给几个数组，第一位没有用，问有没有一个排列能满足这几个数组中元素的先后关系。
数组给出的顺序天然形成有向图。像 1 → 2 1 \rightarrow 2 1→2且 2 → 1 2 \rightarrow 1 2→1这种矛盾的顺序必然是不存在序列的，也就是说给出的关系不能有环。所以简单套一个拓扑排序就可以了
c++
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <time.h>
#include <set>
#include <map>
#include <queue>
#define IOS ios::sync_with_stdio(0);cin.tie(0);
#define rep(index,start,end) for(int index = start;index < end; index ++)
#define drep(index,start,end) for(int index = start;index >= end; index --)
using namespace std;
const int maxn = 2e5+5;
int store[maxn];
vector<int> adj[maxn];
int in_degrad[maxn];
queue<int> Q;
int main() {
IOS
int t;
cin>>t;
while(t--) {
int n,k;
cin>>n>>k;
// init
memset(in_degrad, 0, sizeof(in_degrad));
rep(i,0,n+1) adj[i].clear();
while(!Q.empty()) Q.pop();
rep(i,0,k) {
rep(j,0,n) cin>>store[j];
rep(j,1,n-1) {
int commonA = store[j];
int commonB = store[j+1];
if (find(adj[commonA].begin(), adj[commonA].end(), commonB) == adj[commonA].end()) {
adj[commonA].push_back(commonB);
in_degrad[commonB] ++;
}
}
}
rep(i,1,n+1) {
if (!in_degrad[i])
Q.push(i);
}
while (!Q.empty()) {
int now = Q.front();
Q.pop();
int len = adj[now].size();
rep(i,0,len) {
int next = adj[now][i];
if (-- in_degrad[next] == 0)
Q.push(next);
}
}
bool _loop = false;
rep(i,1,n+1)
if (in_degrad[i]) {
// cout<<i<<' '<<in_degrad[i]<<endl;
_loop = true;
break;
}
cout<<(_loop? "NO":"YES")<<endl;
}
return 0;
}G. One-Dimensional Puzzle 高二排列组合问题
题干太长懒得翻译 有多少种排列方式可以把给出的所有形状拼成一个长条。
大概就是这么个拼接的方法。shape 1和shape 2的个数相差不能超过1，超过就拼不出来；shape 3和shape 4就是造成不同拼接方式的关键，穿插在shape 1和shape 2的间隙，要注意shape 3是可以自拼接，并不是每个间隙只能塞一个
c++
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <time.h>
#include <set>
#include <map>
#include <queue>
#define IOS ios::sync_with_stdio(0);cin.tie(0);
#define mem(A,B) memset(A,B,sizeof(A));
#define rep(index,start,end) for(int index = start;index < end; index ++)
#define drep(index,start,end) for(int index = start;index >= end; index --)
using namespace std;
typedef long long ll;
const int maxn = 3e6+5;
const int mod = 998244353;
ll fact[maxn];
ll pow_mod(ll x, ll p) {
if (p == 0) {
return 1;
}
if (p % 2 == 0) {
ll y = pow_mod(x, p / 2);
return (y * y) % mod;
}
return (x * pow_mod(x, p - 1)) % mod;
}
ll inv(ll x) {
return pow_mod(x, mod - 2);
}
ll cnk(ll n, ll k) {
ll res = fact[n];
res = (res * inv(fact[k])) % mod;
res = (res * inv(fact[n - k])) % mod;
// cout<<"n:"<<n<<" k:"<<k<<" res:"<<res<<endl;
return res;
}
int abs(int num) {
return num<0 ? -num :num;
}
int store[5];
int main() {
IOS
fact[0] = fact[1] = 1;
rep(i,2,maxn)
fact[i] = (fact[i-1] * i) % mod;
int t;
cin>>t;
while(t--) {
rep(i,0,4) cin>>store[i];
if (store[0] == 0 && store[1] == 0) {
cout<<((store[2]!=0 && store[3] != 0)? 0:1)<<endl;
continue;
}
int dfi = abs(store[1] - store[0]);
if (dfi > 1) {
cout<<0<<endl;
continue;
}
ll ans = 0;
if (dfi == 0) {
// same and not 0
int x3,x4;
x3 = store[1];
x4 = x3 + 1;
ans += (cnk(store[2]+x3-1, store[2]) * cnk(store[3]+x4-1, store[3])) % mod;
x4 = store[1];
x3 = x4 + 1;
ans = ans + (cnk(store[2]+x3-1, store[2]) * cnk(store[3]+x4-1, store[3])) % mod;
ans = ans % mod;
} else {
// greater than one
int x3,x4;
x3 = x4 = max(store[0], store[1]);
ans = (cnk(store[2]+x3-1, store[2]) * cnk(store[3]+x4-1, store[3])) % mod;
}
cout<<ans<<endl;
}
return 0;
}