专题三acm1001

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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.<br>

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).<br>

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.<br>

Sample Input

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6
（1）大意：
给出一个数列a[1],a[2],a[3].a[n],我们的工作是计算这个子数列的最大和.例如,给出数列(6,-1,5,4,-7),那么这个数列的最大和就是6 + (-1) + 5 + 4 = 14.
（2）思路：
动态规划的思想，首先求出局部的最大和并保存起来，然后接下来就用本身和前面保存的最大和相比较，选出最大值，到最后就能得到结果。
（3）感想：
思路是有的，但是有思路缺兵也不一定能做对，因为有好多编译错误啥的，关键还是细节，注意容量数组大小，千万就是细节，动态规划千篇一路。
（4）
#include <iostream>#include <cstdio>using namespace std;int main(){    int i,ca=1,t,s,e,n,x,now,before,max;    scanf("%d",&t);    while(t--)    {       scanf("%d",&n);       for(i=1;i<=n;i++)       {         scanf("%d",&now);         if(i==1)         {            max=before=now;            x=s=e=1;         }         else {             if(now>now+before)             {                before=now;                x=i;             }             else before+=now;              }         if(before>max)           {max=before,s=x,e=i;}       }       printf("Case %d:\n%d %d %d\n",ca++,max,s,e);       if(t)    {     printf("\n");    }    }    return 0;}

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