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

 

大意：

一位数组，找出和最大的最长子序列和此序列的开头和结尾

 

思路：

建立一维dp数组，dp[i]的为终点为i的子序列的最大和。

找状态转移规律，

从dp[1]开始， dp[1] = a[1]。

然后 dp[2]，有两种情况，要么是（a[1], a[2]）, 要么是（a[2]），所以dp[2] = max(a[1]+a[2], a[2]), 把a[2]提出来， 就是dp[2] = a[2] +max(a[1], 0), 也就是dp[2] = a[2] + max(dp[1], 0);

然后看dp[3], dp[3] 有三种情况： ( a[1], a[2], a[3])    ( a[2], a[3] )    (a[3]), 所以dp[3]就是这三个数中最大的， 前两个数最大其实就是dp[2]，所以最后是 dp[3] = a[3] + max(dp[2], 0);

那么dp[i] = a[i] + max(dp[i-1], 0)

 

#include<cstdio> #include<cstdlib> #include<cmath> #include<cstring> #include<string> #include<iostream> #include<stack> #include<queue> #include<algorithm> #include<map> #include<list> #include<vector> #define INF 0x3f3f3f3f #define MAXN 100005 using namespace std; typedef struct { int sum; int l; int r; }node; int main() { int T, Case = 1; scanf("%d", &T); while(T--) { int n, a[MAXN], i, j; node dp[MAXN] = {0}; scanf("%d", &n); for(i = 1; i <= n; i++) { scanf("%d", &a[i]); } dp[1].sum = a[1]; dp[1].l = 1; dp[1].r = 1; for(i = 2; i <= n; i++) { if(dp[i-1].sum >= 0) { dp[i].sum = a[i] + dp[i-1].sum; dp[i].l = dp[i-1].l; dp[i].r = i; } else { dp[i].sum = a[i]; dp[i].l = i; dp[i].r = i; } } int Max = -INF, maxi; for(i = 1; i <= n; i++) { if(dp[i].sum > Max) { Max = dp[i].sum; maxi = i; } } printf("Case %d:\n", Case++); printf("%d %d %d\n", dp[maxi].sum, dp[maxi].l, dp[maxi].r); if(T != 0) printf("\n"); } return 0; }