动态规划，区间最大和

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 183319    Accepted Submission(s): 42766

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

 

动态规划问题，给你一个数组，让你求从一个从i到j和最大的这么一个范围，策略是：每输入第i个数要去判断要不要把这个数加入我前面所组成的数组x[i-1]里，
如果加入这个数后得到和不小于这个数，那就把这个数加入前面的数列，否则重新开始构建最大和的数列。
用两个数组x[i]存max(x[i]+a,a),x1[i]存处理每一个数时开始的位置
例子：输入 5   6 -1 5 4 7时

	6	-1	5	4	-7
x	6	5	10	14	7
x1	1	1	1	1	1

循环一遍找出最大的和为14，开始的位置为1，结束位置为4（第4个数）
例子输入 8  0 6 -7 1 6 1 -5 9 （自己造的例子）

	0	6	-7	1	6	1	-5	9
x	0	6	-1	1	7	8	3	12
x1	1	1	1	4	4	4	4	4

（输入第4个数的时候（-1+1=0<1）所以重新从第四个数开始）
循环一遍找出最大的和为12，开始的位置为4，结束位置为8（第8个数）

    #include<iostream>  
    #include<cstdio>  
    #include<cstring>  
    using namespace std;  
    const int INF=0x3f3f3f3f;  
    int main()  
    {  
        int T;  
        int x[100005];  
        int x1[100005];  
        int cas=1;  
        scanf("%d",&T);  
        while(T--)  
        {  
            memset(x,0,sizeof(x));  
            memset(x1,0,sizeof(x1));  
            //memset(x2,0,sizeof(x2));  
            int a,b=-INF,n;  
            scanf("%d",&n);  
            x[0]=-INF;  
            x1[0]=1;  
            for(int i=1;i<=n;i++)  
            {  
                scanf("%d",&a);  
                if(x[i-1]+a>=a)  
                {  
                    x[i]=x[i-1]+a;  
                    x1[i]=x1[i-1];  
                }  
                //x[i]=max(x[i-1]+a,a);  
                else  
                {  
                    x[i]=a;  
                    x1[i]=i;  
                }  
            }  
            int s,e;  
            for(int i=1;i<=n;i++)  
            {  
                if(x[i]>b)  
                {  
                    b=x[i];  
                    s=x1[i];  
                    e=i;  
                }  
            }  
            printf("Case %d:\n",cas++);  
            printf("%d %d %d\n",b,s,e);  
            if(T!=0)printf("\n");  
        }  
    }