最长数字和的自串

HDU1003(http://acm.hdu.edu.cn/showproblem.php?pid=1003)

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 

////////////////////////////////////////////////////////// 

求最大和子串，用动态规划的方法来做。令f[n]为到第n个位置时的最大和。则f[n]=max{f[n-1]+data[n],data[n]}即如果f[n-1]+data[n]>data[n],则f[n]=f[n-1]+data.也就是f[n-1]>0。如果f[n-1]>0,则后一项加上f[n-1]肯定更大。所以本题主要求f[1..n]。然后求出f中最大的一个（max(f[1..n])）。这题又要求最大串的开始与结束位置，当f[n-1]<0时，开始位置要重新开始记录。主要思想是这样，编写的时候可以换一种实现形式。如下 

#include<stdio.h>int main(){ int s,e,max,sum,n,t,dat,i,s1,j; for (;scanf("%d",&t)!=EOF;) {  j=1;  while (t--)  {   max=-10000;sum=0;   scanf("%d",&n);   s1=1;   for(i=1;i<=n;i++)   {    scanf("%d",&dat);    sum+=dat;    if(sum>max)    {     max=sum;s=s1;e=i;    }    if(sum<0)    {     sum=0;s1=i+1;    }   }   if(j-1!=0) printf("\n");   printf("Case %d:\n",j++);   printf("%d %d %d\n",max,s,e);     } 

 } return 0;} 

包括串全部为负数的情况。 

通过保留上一个最大值数字和开始结束位置。当遇到更大时对原数据进行替换