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
Author
Ignatius.L
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很简单的dp,dp[i]表示以i为开头的值
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn = 100010; const int inf = 0x3f3f3f3f; int a[maxn],dp[maxn]; int T,n; 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]); dp[i] = a[i]; } for(int i = n-1; i >= 1; i--){ dp[i] = max(dp[i], dp[i+1] + a[i]); } // for(int i = 1; i <= n; i++) // printf("%d ",dp[i]); int index; int max1 = -inf; for(int i = 1; i <= n ;i++){ if(max1 < dp[i]){ max1 = dp[i]; index = i; } } int res = 0; int index1; for(int i = index; i <= n; i++){ res += a[i]; if(res == max1){ index1 = i; break; } } printf("Case %d:\n",cas); printf("%d %d %d\n",max1,index,index1); if(cas < T) printf("\n"); } return 0; }