Problem - 466C - Codeforces C. Number of Ways

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#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;long long a[100217*5];long long ani;int c[100217*5];int d[100217*5];int main(){    int n,m;    while(scanf("%d",&n)!=-1)    {        ani=0;        memset(a,0,sizeof(a));        for(int i=1;i<=n;i++)        {            int temp=0;            scanf("%d",&temp);            a[i]=a[i-1]+temp;        }        if(a[n]%3!=0)        {            puts("0");        }        else        {           long long x=a[n]/3;           long long y=a[n]/3*2;           int b[2]={0,0};           int t=0;           long long sum=0;           for(int i=1;i<n-1;i++)            if(a[i]==x)            {                c[b[0]]=i;//存储值为x的下标                b[0]++;            }           for(int i=1;i<=n-1;i++)           if(a[i]==y)           {               d[b[1]]=i;//<span style="font-family: Arial, Helvetica, sans-serif;">存储值为y的下标</span>               b[1]++;           }           for(int i=0;i<b[0];i++)           {               t=upper_bound(d,d+b[1],c[i])-d;//查找有哪些y在x的后方               sum+=(long long)(b[1]-t);           }           printf("%I64d\n",sum);        }    }}

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