最大字串和



Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 204790 Accepted Submission(s): 47879



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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).

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

[align=left]Sample Output[/align]

Case 1:
14 1 4

Case 2:
7 1 6

#include<iostream>

#include<cstdio>

using namespace std;

int main()

{

int j,i,k,n,m,t;

int a; //不需要数组，只需要一个输入变量

scanf("%d",&t);

for (j=1;j<=t;j++)

{

scanf("%d",&n);

int sum=0,maxsum=-10000002,first =0, last = 0, temp = 1;

for (i=0;i<n;i++)

{

scanf("%d",&a);

sum += a;

if (sum > maxsum)

{

maxsum = sum;first = temp;last = i+1;

}

if (sum < 0)

{

sum = 0;temp = i+2;

}

}

printf("Case %d:\n%d %d %d\n",j,maxsum,first,last);

if (j!=t)

{

printf("\n");

}

}

return 0;

}

最大字串和：没有输出最大字串和的未知的代码：

#include<iostream>

#include<cstdio>

using namespace std;

int main()

{

int j,i,k,n,m,t;

int a; //不需要数组，只需要一个输入变量

scanf("%d",&t);

for (j=1;j<=t;j++)

{

scanf("%d",&n);

int sum=0,maxsum=-10000002;

for (i=0;i<n;i++)

{

scanf("%d",&a);

sum += a;

if (sum > maxsum)

{

maxsum = sum;;

}

if (sum < 0)

{

sum = 0;;

}

}

printf("Case %d:\n%d \n",j,maxsum);

if (j!=t)

{

printf("\n");

}

}

return 0;

}