Max Sum
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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
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Sample Output
Case 1:
14 1 4
Case 2:
7 1 6
经典DP 不解释!
代码如下:
#include<cstdio>int main(){ int num[100000]; int n,t,cur,before,max,s,e,ca=1; scanf("%d",&n); while(n--) { scanf("%d",&t); for(int i=1;i<=t;i++) scanf("%d",&num[i]); max=before=num[1]; s=e=cur=1; for(int i=2;i<=t;i++) { if(before<0) { before=num[i]; cur=i; } else before+=num[i]; if(max<before) { max=before; s=cur; e=i; } } printf("Case %d:\n%d %d %d\n",ca++,max,s,e); if(n) printf("\n"); } return 0;}
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