hdu1003 最大连续子序和

最新推荐文章于 2020-11-03 20:32:45 发布

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本文介绍了一种寻找整数序列中具有最大和的子序列的算法，并提供了两种不同的实现方式。通过输入一系列整数，算法能找出这些整数构成的最大和及对应的起始和结束位置，适用于解决特定类型的问题。

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52833 Accepted Submission(s): 11726

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

#include <stdio.h> int main() {int n,T,a,sta,end,max,k,i,p,t; scanf("%d",&T); for(p=1;p<=T;p++) { scanf("%d",&n); max=-9999; //因为一个数a 是-1000~1000的，所以这里相当于变成最小值 t=0; //表示某段连续和 sta=end=k=1; // sta最大和的开始,end最大和的结束,k记录每次求和的开始 for(i=1;i<=n;i++) { scanf("%d",&a); t+=a; if(t>max) { //记录最大连续和的值 max=t; sta=k; end=i; } if(t<0) { t=0; k=i+1; } } if(p!=1) printf("/n"); printf("Case %d:/n",p); printf("%d %d %d/n",max,sta,end); } }

下面是用数组做的：

#include <stdio.h> #include <string.h> int a[100002]; int t,n,max,sum,l,r,i,j,k; int main() { scanf("%d",&t); for(i=1;i<=t;i++){ scanf("%d",&n); bool pd=0; sum=max=l=r=0; k=1; for(j=1;j<=n;j++) { scanf("%d",a+j); if(a[j]>0) pd=1; sum+=a[j]; if(sum<0) sum=0,k=j+1; if(sum>max) max=sum,l=k,r=j; } if(!pd){ max=a[1],l=1,r=1; for(j=2;j<=n;j++) if(a[j]>max) max=a[j],l=j,r=j; } printf("Case %d:/n",i); printf("%d %d %d/n",max,l,r); if(i<t) puts(""); } }