NOIP2011 D2 T2 聪明的质检员

这题目有毒

花了一个下午在这水(shen)题上



，最后才看见0
< S≤10^12。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<cmath>
#pragma warning(disable:4996)
using namespace std;
#define MAXN 200009
#define LL long long

LL n , m ;
LL S;
struct _
{
LL val , wei;
}a[MAXN];
struct __
{
int l , r;
}b[MAXN];
LL c[MAXN];//前缀和数组
LL d[MAXN];//前缀和，w[j]>W个数
LL L , R , cnt;
void count(LL L, LL R)
{
if (L == R)
return;
LL mid = (L + R) >> 1;
memset(c , 0 , sizeof(c));
memset(d , 0 , sizeof(d));
for (int i = 1; i <= n; i++)
{
if (a[i].wei >= mid)
{
c[i] = c[i - 1] + a[i].val;
d[i] = d[i - 1] + 1;
}
else
{
c[i] = c[i - 1];
d[i] = d[i - 1];
}
}

LL ans = 0;
for (int i = 1; i <= m; i++)
{
ans += ((c[b[i].r] - c[b[i].l - 1])*(d[b[i].r] - d[b[i].l - 1]));
}
if (abs(S - ans) < abs(S - cnt)) cnt = ans;
if (S > ans)
count(L , mid);
if (S == ans)
{
cnt = ans;
return;
}
if (S < ans)
count(mid + 1 , R);
}

int main()
{
scanf("%lld%lld%lld" , &n , &m , &S);
for (int i = 1; i <= n; i++)
{
scanf("%lld%lld" , &a[i].wei , &a[i].val);
}
for (int i = 1; i <= m; i++)
scanf("%lld%lld" , &b[i].l , &b[i].r);

count(1 , 100000);
printf("%lld\n" , abs(cnt - S));
return 0;
}