Max Sum
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Max Sum
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
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Sample Output
14 1 4
Case 2:
7 1 6
#include <stdio.h>#include <string.h>int dp[50005];int main(){ int T; scanf("%d",&T); for(int u=1; u<=T; u++) { int m; scanf("%d",&m); for(int i=1; i<=m; i++) scanf("%d",&dp[i]); int sum =-1000,ans =0,si = 1,ei,ej; for(int i=1; i<=m; i++) { ans += dp[i]; if(ans>sum) { sum = ans; ei = si; ej = i; } if(ans <0) { ans = 0; si = i+1; } } printf("Case %d:\n",u); printf("%d %d %d\n",sum,ei,ej); if(u!=T) printf("\n"); } return 0;}
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