OpenJudge_P1481 Maximum sum(最大双子序列和)

总时间限制: 1000ms 内存限制: 65536kB
描述
Given a set of n integers: A={a1, a2,…, an}, we define a function d(A) as below:

Your task is to calculate d(A).
输入
The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, …, an. (|ai| <= 10000).There is an empty line after each case.
输出
Print exactly one line for each test case. The line should contain the integer d(A).
样例输入
1

10
1 -1 2 2 3 -3 4 -4 5 -5
样例输出
13
提示
In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.

Huge input,scanf is recommended.
来源
POJ Contest,Author:Mathematica@ZSU

第一眼看去就是初赛题
m计算没有初始化成INT_MIN，初始化成0，无限WA

#include<cstdio>#include<cstring>#include<climits>#include<iostream>using namespace std;#define N 50005int t,n,ans,temp,m;int a[N],f1[N],f2[N];int main(){    scanf("%d",&t);    while(t--){        memset(f1,0,sizeof(f1));memset(f2,0,sizeof(f2));        scanf("%d",&n);temp=0,m=ans=INT_MIN;        for(int i=1;i<=n;i++) scanf("%d",&a[i]);        for(int i=1;i<=n;i++){            temp+=a[i];if(temp>m) m=temp;             if(temp<0) temp=0;            f1[i]=m;        }        temp=0,m=INT_MIN;        for(int i=n;i>=1;i--){            temp+=a[i];if(temp>m) m=temp;             if(temp<0) temp=0;            f2[i]=m;        }        for(int i=1;i<n;i++){            ans=max(ans,f1[i]+f2[i+1]);        }        printf("%d\n",ans);    }    return 0;}