HDOJ1003Max Sum

Problem Description

Given a sequence a[1],a[2],a[3]……a
, 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

有2种方法做，我开始不明白题目意思，走了很多弯路。

题目要求是这样的：

输出数组的子序列的最大值。

如果有相同的，开始序列输出小的。

结尾序列输出大的。

/**1003题**/
#include <stdio.h>
#include <stdlib.h>
int a[1000020];
long long int c[3000020],b[1000020];
int main()
{
int t,tt=1;
scanf("%d",&t);
while(t--)
{
int i,j,n,a1,a2;
scanf("%d",&n);
int age=0;
if(n!=1)
{
for(i=0; i<n; i++)
scanf("%d",&a[i]);
long long int sum=0;
for(i=0; i<n; i++)
{
if(a[i]>=0)
{
a1=a2=i+1;
age=1;
break;
}
}
if(age!=1)
{
int max=a[0];
a1=a2=1;
for(i=1; i<n; i++)
{
if(a[i]>max)
{
max=a[i];
a1=a2=i+1;
}
}
printf("Case %d:\n",tt);
tt++;
printf("%d %d %d\n",max,a1,a2);
if(t!=0)
printf("\n");
}
else
{
int k=0;
c[k]=0;
long long int max=c[k];
for(i=a1-1;i<n;i++)
{
c[k]=0;
for(j=i;j<n;j++)
{
c[k]=a[j]+c[k];
if(c[k]<0)
break;
if(max<c[k])
{
a1=i+1;
a2=j+1;
max=c[k];
}
if(max==c[k])
{
if(a1>i+1)
a1=i+1;
if(a2<j+1)
a2=j+1;
}
}
k++;
}
printf("Case %d:\n",tt);
tt++;
printf("%I64d %d %d\n",max,a1,a2);
if(t!=0)
printf("\n");
}
}
if(n==1)
{
scanf("%d",&a[0]);
{
printf("Case %d:\n",tt);
tt++;
printf("%d 1 1\n",a[0]);
if(t!=0)
printf("\n");
}
}
}
return 0;
}

第二种方法：

#include<stdio.h>
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)//初始化
{
max=before=now;
//max保留之前算出来的最大和，before存储目前在读入数据前保留的和，now保留读入数据
x=s=e=1;
//x用来暂时存储before保留的和的起始位置，
//当before>max时将赋在s位置，s，e保留最大和的start和end位置
}
else
{
if(now>now+before)
//如果之前存储的和加上现在的数据比现在的数据小，
//就把存储的和换成现在的数据，反之就说明数据在递增，可以直接加上
{
before=now;
//预存的位置要重置
}
else before+=now;
}
if(before>max)
//跟之前算出来的最大和进行比较，如果大于，位置和数据就要重置
max=before,s=x,e=i;
}
printf("Case %d:\n%d %d %d\n",ca++,max,s,e);
if(t)printf("\n");
}
return 0;
}

#include <cstdio>
using namespace std;

int T,n,m;
int post1,post2,x;//post1表示序列起点，post2表示序列终点，x表示每次更新的起点
int max,now;//max表示最大子序列和，now表示各个子序列的和
int i,j;

int main ()
{
scanf ("%d",&T);
for (i=1; i<=T; i++)
{
scanf ("%d%d",&n,&m);
max = now = m;
post1 = post2 = x = 1;//初始化
for (j=2; j<=n; j++)
{
scanf ("%d",&m);
if (now + m < m)//对于每个数，如果该数加上当前序列和比本身还小
{
now = m;//更新区间
x = j;//更新起点
}
else
now += m;//否则把该数加进序列
if (now > max)//如果当前序列和比已有最大序列和大，更新
{
max = now;
post1 = x;//记录新的起点和终点
post2 = j;
}
}
printf ("Case %d:\n",i);
printf ("%d %d %d\n",max, post1, post2);
if (i != T)
printf ("\n");
}
return 0;
}