HDU1003 最大连续子序列

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 324393    Accepted Submission(s): 77146

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

 

 

【题意】

　　给定一个序列a[1]，a[2]，a[3]，.....，a[n]，来计算子序列的最大和。例如，给定序列(6，-1，5，4，-7)，这个序列中的最大和是 6 +(-1)+ 5 + 4 = 14。

　　输入的第一行包含一个整数T(1<=T<=20)，它表示测试用例的数量。然后是T行，每一行以N开头(1<=N<=100000)，然后是N个整数(所有整数都在-1000到1000之间)

　　对于每个测试用例，输出两行。第一行是“Case #:”，#表示测试用例的数量。第二行包含三个整数，序列中的最大和，子序列的起始位置，子序列的结束位置。如果有多个结果，输出第一个。在两种情况之间输出空行。

【代码】

#include <stdio.h>
#include <stdlib.h>
int main(){
    int i,j,n,t;
    scanf("%d",&t);
    for(i=1;i<=t;i++){
        int *a,first=0,last=0,sum=0,tmp=1,max=-1001;
        scanf("%d",&n);
        a=(int*)malloc(n*sizeof(int));
        for(j=0;j<n;j++){
            scanf("%d",&a[j]);
            sum += a[j];
            if(sum>max){
                max=sum;
                first=tmp;
                last=j+1;
            }
            if(sum<0){
                sum=0;
                tmp=j+2;
            }
        }
        printf("Case %d:\n%d %d %d\n",i,max,first,last);
        if(i!=t)printf("\n");
        a=NULL;
    }
    return 0;
}