AtCoder Beginner Contest 067 C

C - Splitting Pile
Time limit : 2sec / Memory limit : 256MB
Score : 300 points
Problem Statement
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer aiwritten on it.
They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card.
Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x−y|. Find the minimum possible value of |x−y|.
Constraints
2≤N≤2×105
−109≤ai≤109
ai is an integer.
Input
Input is given from Standard Input in the following format:
Na1 a2 … aN
Output
Print the answer.
Sample Input 1
Copy
61 2 3 4 5 6
Sample Output 1
Copy
1
If Snuke takes four cards from the top, and Raccoon takes the remaining two cards,x=10, y=11, and thus |x−y|=1. This is the minimum possible value.
Sample Input 2
Copy
210 -10
Sample Output 2
Copy
20
Snuke can only take one card from the top, and Raccoon can only take the remaining one card. In this case, x=10, y=−10, and thus |x−y|=20.

题解：前缀和查询

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <cmath>#include <vector>#include <algorithm>using namespace std;#define max(x,y) (x>y?x:y)#define min(x,y) (x<y?x:y)#define MAX 100000000000000000#define MOD 1000000007#define INF 1000000000#define mem(a) (memset(a,0,sizeof(a)))typedef long long ll;ll x,sum[200009],n,ans,minn;int main(){    scanf("%lld",&n);    sum[0]=0;    for(ll i=1;i<=n;i++)    {        scanf("%lld",&x);        sum[i]=sum[i-1]+x;    }    ans=sum[n];    minn=abs(2*sum[1]-sum[n]);    for(ll i=2;i<=n-1;i++)    {        minn=min(minn,abs(2*sum[i]-sum[n]));    }    printf("%lld\n",minn);    return 0;}