2022-2023 ICPC Brazil Subregional Programming Contest VP记录

2022-2023 ICPC Brazil Subregional Programming Contest VP SerialABCDEFGHIJKLMNSolve●●●●●●◉●●●○○○● A.Finding Maximal Non-Trivial Monotones

找连续且长度大于等于a的子串长度和。直接记录加和即可。

#include 
#define int long long
#define endl '\n'
using namespace std;

const int N = 1e5 + 10;
int a[N];

inline void solve(){
    int n = 0; cin >> n;
    string s; cin >> s, s = '@' + s;
    int cnt = 0, ans = 0;
    for(int i = 1; i = 2) ans += cnt;
            cnt = 0;
        }
    }
    if(cnt >= 2) ans += cnt;
    cout =1,(x*=x)%=mod) if(y&1) (res*=x)%=mod;
    return res;
}
int ck(int x,int y,int z){
    int res=0;
    for(int i=0;i>i)&1),fg2=((y>>i)&1),fg3=((z>>i)&1);
        if(fg1!=fg2){
            res+=(fg1>i)&1);
        int now=fg1*4+fg2*2+fg3;
        for(auto t:way){
            for(int j=0;j>l1>>r1>>l2>>r2>>l3>>r3;
    int ans1=dfs(r1,r2,r3),
        ans2=dfs(r1,r2,l3-1),
        ans3=dfs(r1,l2-1,r3),
        ans4=dfs(r1,l2-1,l3-1),
        ans5=dfs(l1-1,r2,r3),
        ans6=dfs(l1-1,r2,l3-1),
        ans7=dfs(l1-1,l2-1,r3),
        ans8=dfs(l1-1,l2-1,l3-1);
    int fenzi=((ans1-ans2-ans3+ans4-ans5+ans6+ans7-ans8)%mod+mod)%mod;
    int fenmu=(r1-l1+1)*(r2-l2+1)%mod*(r3-l3+1)%mod;
    cout p[i].y;
    int lastx = 0, lasty = 0;
    for(int i = 1; i