HDU-1003-Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 113689    Accepted Submission(s): 26291

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

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

 

Sample Output

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

 

Author

Ignatius.L

 

import java.util.Scanner;public class HDU_1003_Max_Sum {/** * Accepted1003421MS3636K966 BJava1983210400 */public static void main(String[] args) {         int max,sum,A,B,a,b,t;  Scanner scan=new Scanner(System.in);   int N=scan.nextInt();    for(int i=1;i<=N;i++){                 int M = scan.nextInt();   A=B=a=b=1;   max=sum=-1001;;    for(int j= 1; j<=M ; j++){        t=scan.nextInt();if(sum + t < t){    sum = t; a=b=j; //每一次读入t时与sum+t,若小于读入的值，换个位置，再从新读取}         else {sum += t ; ++b;//若大于读入的值，正好，那就最大连续串和序列的末尾位置b需+1，即(++b)，同时将新的成员加入sum         }       if(max<sum){max=sum;A=a;B=b;//大于max时再改变其值，起始位置，终止位置}   }System.out.println("Case "+i+":");//换行，循序渐进！ System.out.println(max+" "+A+" "+B);if(i!=N)  System.out.println();   /*System.out.println("Case "+i+":\n"+max+" "+A+" "+B);//PE,这样的格式输出不对！！切记！，虽然看着对！   if(i!=N)   System.out.println();*/ }}}

C语言版：

#include <stdio.h>   int main()   {       int z,n,max,sum;       int a,b,A,B,t;       scanf("%d",&z);       for(int k=1;k<=z;k++)       {           scanf("%d",&n);           sum = max = -1001;           for(int i=1;i<=n;i++)           {               scanf("%d",&t);               if(sum+t < t)                   sum = t , a = b = i;      //a、b记录当前连续子序列的起始、结束位置               else                  sum += t , ++b;               if(max < sum)                   max = sum , A = a , B = b;           }           printf("Case %d:\n%d %d %d\n",k,max,A,B);           if(k-z) puts("");       }       return 0;   }