51NOD1495 中國好區間 【尺取法】

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尺取法求[i,j]區間,使得[i,j]區間size>=k且>=T的數恰好有k個,於是[1,j],[2,j],[3,j],….,[i,j]都滿足題意,以j結尾有i種方案

#include<stdio.h>
#include<bits/stdc++.h>
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define MEM(a,x) memset(a,x,sizeof(a))
#define lowbit(x) ((x)&-(x))

using namespace std;

const int inf=1e9+7;
const int N = 1e7+5;

int num;//>=T的數個數

ll a[N];

ll slove(int n,int T,int k){
    num=0;
    int i,j;
    for(i=1;num<k&&i<=n;++i){
        num+=(a[i]>=T);
    }
    if(num<k){
        return 0;
    }
    ll ans=0;
    j=i-1;
    i=1;
    for(;j<=n;++j){
        while((j-i+1>=k+1)&&(a[i]<T||(num-1>=k))){
            num-=(a[i]>=T);
            ++i;
        }
        ans+=i;
        num+=(a[j+1]>=T);
    }
    return ans;
}

int main(){
    //freopen("/home/lu/code/r.txt","r",stdin);
    //freopen("/home/lu/code/w.txt","w",stdout);
    int n,k,T,a0,b,c,p;
    while(~scanf("%d%d%d%d%d%d%d",&n,&k,&T,&a0,&b,&c,&p)){
        a[0]=a0;
        for(int i=1;i<=n;++i){
            a[i]=(a[i-1]*b+c)%p;
        }
        printf("%lld\n",slove(n,T,k));
    }
    return 0;
}