Max Sum(简单的最大子序列和的问题)

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A - Max Sum

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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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

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

Sample Output

Case 1:14 1 4Case 2:7 1 6

题目大意：求所给序列的最大子序列之和，并求出该最大子序列的首位和末位的位置。

分析：开一个dp数组记录以每位数字结尾的最大子序列之和，比较求出最大的子序列之和，并记录该子序列结尾的

数字序号，从结尾数字向前累加，直到和等于求得的最大子序列之和，记录此时的数字序号，便是该最大子

序列的首位。

注意：1，在第二个样例中，0作为最大子序列的首位。

2，当最大子序列为一个数字时，首位与末位相同。

3，最后一组结果后不需要空行。

代码如下：

#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<cstdio>
using namespace std;
int a[100005],dp[100005];
int main()
{
int i,j,k,n,t,mm,nn,st,en,sum;
scanf("%d",&t);
for(i=0;i<t;i++)
{
sum=0;
memset(dp,0,sizeof(dp));
scanf("%d",&n);
for(j=1;j<=n;j++)
scanf("%d",&a[j]);
dp[1]=a[1];
for(j=2;j<=n;j++)
{
if(a[j]+dp[j-1]>a[j])
dp[j]=a[j]+dp[j-1];
else
dp[j]=a[j];
}
mm=dp[1];
nn=1;
for(j=2;j<=n;j++)
{
if(dp[j]>mm)
{mm=dp[j];nn=j;}
}
if(mm==a[nn]&&(dp[nn-1]!=0||nn-1<1))
{st=nn;en=nn;}
else
{
en=nn;
sum=a[nn];
for(j=nn-1;j>=1;j--)
{
sum+=a[j];
if(sum==mm)
{
if(a[j-1]==0&&j-1>=1)
st=j-1;
else
st=j;
break;
}
}
}
printf("Case %d:\n",i+1);
printf("%d %d %d\n",mm,st,en);
if(t-i>1)
printf("\n");
}
getchar();
getchar();
return 0;
}

ps：第一次写博客，有错误望大家指出

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