HDU 1003：Max Sum（DP）

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 142742    Accepted Submission(s): 33225

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

 

Sample Input

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

这里再给出一些测试数据：

4 0 0 2 0 —— 2 1 36 2 7 -9 5 4 3 —— 12 1 64 0 0 -1 0 —— 0 1 17 -1 -2 -3 -2 -5 -1 -2 —— -1 1 16 -1 -2 -3 1 2 3 —— 6 4 6

5 -3 -2 -1 -2 -3 —— -1 3 3

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<stdlib.h>#include<vector>#include<queue>#include<cmath>using namespace std;const int maxn = 100000 + 150;int t;int n;int temp;int start;int last;int a[maxn];int ans;int sum;int main(){    scanf("%d", &t);    for(int cas=1; cas<=t; cas++)    {        scanf("%d", &n);        for(int i=1; i<=n; i++)            scanf("%d", &a[i]);        sum = a[1];        ans = a[1];//开始将其赋值为0 。。WA。。        temp = 1;        start = 1;        last = 1;//以为last不用初始也可以。。WA。。        for(int j=2; j<=n; j++)        {            if( sum<0 )            {                temp = j;                sum = 0;            }            sum = sum + a[j];            if( sum>ans )            {                ans = sum;                last = j;                start = temp;            }        }        printf("Case %d:\n", cas);        printf("%d %d %d\n", ans, start, last);        if( cas<t )            printf("\n");    }    return 0;}