Two half board

To P1314 smart quality supervision , for example
the first writing:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
const int N=200000+10;
int v[N],w[N],n,m;
int l[N],r[N];
int sum[N],cnt[N];
LL f(int W)
{
    memset(sum,0,sizeof(sum));
    memset(cnt,0,sizeof(cnt));
    for(int i=1;i<=n;i++)
    {
        if(w[i]>=W)
        {
            cnt[i]=cnt[i-1]+1;
            sum[i]=sum[i-1]+v[i];
        }
        else
        {
            cnt[i]=cnt[i-1];
            sum[i]=sum[i-1];
        }
    }
    LL res=0;
    for(int i=1;i<=m;i++)
    {
        res=res+(sum[r[i]]-sum[l[i]-1])*(cnt[r[i]]-cnt[l[i]-1]);
    }
    return res;
}
int main()
{
    LL S;
    cin>>n>>m>>S;
    for(int i=1;i<=n;i++)
        scanf("%d%d",w+i,v+i);
    for(int i=1;i<=m;i++)
         scanf("%d%d",l+i,r+i);
    int L=0,R=1000000+1,mid;
    the while (L <R & lt)                    // where demand is a function of the value of a decreasing function of the abscissa is strictly less than the minimum point S 
    { 
        MID = (L + R & lt) / 2 ;       // the rounding board 
        IF (F ( mid) < S) 
            R = mid;       // mid of the function value is less than S, can not be guaranteed at the mid-1 function value is less than S, R can be reduced to mid 
        the else 
            L = mid + . 1 ;       // function value is greater than or equal at the mid S, L can be directly reduced to make the final mid + 1 L and R will be equal, not infinite loop 
    }                          
    COUT << min (Sf of (L), F (L- . 1 ) -S) << endl;
     return  0 ; 
}

The second writing:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
const int N=200000+10;
int v[N],w[N],n,m;
int l[N],r[N];
int sum[N],cnt[N];
LL f(int W)
{
    memset(sum,0,sizeof(sum));
    memset(cnt,0,sizeof(cnt));
    for(int i=1;i<=n;i++)
    {
        if(w[i]>=W)
        {
            cnt[i]=cnt[i-1]+1;
            sum[i]=sum[i-1]+v[i];
        }
        else
        {
            cnt[i]=cnt[i-1];
            sum[i]=sum[i-1];
        }
    }
    LL res=0;
    for(int i=1;i<=m;i++)
    {
        res=res+(sum[r[i]]-sum[l[i]-1])*(cnt[r[i]]-cnt[l[i]-1]);
    }
    return res;
}
int main()
{
    LL S;
    cin>>n>>m>>S;
    for(int i=1;i<=n;i++)
        scanf("%d%d",w+i,v+i);
    for(int i=1;i<=m;i++)
         scanf("%d%d",l+i,r+i);
    int L=0,R=1000000+1,mid;
    the while (L <R & lt)                    // where demand is a function of the value of a decreasing function exceeds the maximum of the abscissa of the point S 
    { 
        MID = (L + R & lt + . 1 ) / 2 ;       // the rounding board 
        IF (F (mid)> = S) 
            L = mid;       // function value is greater than or equal at mid S, can not be guaranteed at the mid + 1 is greater than the value of the function equal to S, L can be reduced to mid 
        the else 
            R & lt = mid- 1 ;       // mid the function value is less than S, R can be directly reduced to make the final mid-1 L and R will be equal, not infinite loop 
    }                          
    COUT << min (F (L) -S, Sf of (L + . 1 )) << endl;
     return  0 ; 
}