hdu1003Max Sum

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

最大子段和，dp

代码如下：

#include <cstdio>#include <iostream>#include <cstring>#include <cctype>#include <cmath>#include <algorithm>#include <queue>#include <stack>#include <set>#include <map>#include <cstdlib>#include <vector>#define ll long longusing namespace std;int main(){    int i, j, k, t, n, sum, a, b, x, y, z, max;    scanf( "%d", &t);    for ( i = 1; i <= t; i++){        scanf( "%d", &n);        scanf( "%d", &sum);        max = sum;        a = b = x = y = 1;        for ( j = 2; j <= n; j++){            scanf( "%d", &z);            if ( sum >= 0){                sum = sum + z;                y = j;            }            else{                sum = z;                x = y = j;            }            if ( sum > max){                max = sum;                a = x;                b = y;            }        }        printf( "Case %d:\n", i);        printf( "%d %d %d\n", max, a, b);         if ( i != t)        printf( "\n");    }    return 0;}