[Luogu] 聪明的质监员

https://www.luogu.org/problemnew/show/P1314

满足单调性

所以，二分W，进行检验

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;
const int N = 2e5 + ;

#define LL long long
#define oo 999999999999
#define gc getchar()

int n, m;
LL s;
int W[N], V[N], L[N], R[N];
int sum[N], tot[N];
LL Answer = oo;

inline LL read() {
    LL x = ; char c = gc;
    while(c < '' || c > '') c = gc;
    while(c >= '' && c <= '') x = x *  + c - '', c = gc;
    return x;
}

bool see(int w) {
    sum[] = tot[] = ;
    for(int i = ; i <= n; i ++) {
        sum[i] = sum[i - ];
        tot[i] = tot[i - ];
        if(W[i] >= w) sum[i] += V[i], ++ tot[i];
    }
    LL ans();
    for(int i = ; i <= m; i ++) ans += (sum[R[i]] - sum[L[i] - ]) * (tot[R[i]] - tot[L[i] - ]);
    ans = ans - s;
    Answer = min(abs(ans), Answer);
    return ans > ;
}

int main() {
    n = read(), m = read(), s = read();
    for(int i = ; i <= n; i ++) W[i] = read(), V[i] = read();
    for(int i = ; i <= m; i ++) L[i] = read(), R[i] = read();

    LL l = , r = oo;
    while(l <= r) {
        LL mid = (l + r) >> ;
        if(see(mid)) l = mid + ;//比标准值大
        else r = mid - ;
    }
    cout << Answer;

    return ;
}
/*
5 3 15
1 5
2 5
3 5
4 5
5 5
1 5
2 4
3 3
*/