Max Sum
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Given a sequence a1 1,a2 2,a3 3......an 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 sequen
ce, 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 Sample Output
ce, 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.
25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5
Case 1:14 1 4Case 2:7 1 6
#include<stdio.h>#include<string.h>#include<math.h>int a[100010];int main(){ int t; scanf("%d",&t); int g=0; while(t--) { ++g; int n; scanf("%d",&n); memset(a,0,sizeof(a)); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } int sum=0; printf("Case %d:\n",g); int p=1,l=1,k=1; int maxn=-1e9; for(int i=1;i<=n;i++) { sum+=a[i]; if(maxn<sum) { maxn=sum; p=i;//记序列的右边位置 k=l;//记序列的左边位置 } if(sum<0) { l=i+1; sum=0; } } printf("%d %d %d\n",maxn,k,p); if(t) printf("\n"); //} }}
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