hdu1003 Max Sum (求连续子区间最大和)
2015-05-14 20:22
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Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 169481 Accepted Submission(s): 39557
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
Author
Ignatius.L
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解析:求连续子区间最大和。
用f【i】表示以第 i 个数字结尾的连续子区间的最大值,则 f【i】=max{f【i-1】+a【i】,a【i】},然后分情况处理一下起点、终点,并更新当前最优值即可。
代码:
#include<cstdio> #define inf 1000000000 #define maxn 1000 using namespace std; int ans,ans_l,ans_r; void work() { int i,j,k,n,x,sum,l,r; scanf("%d",&n); ans=-inf,sum=0,l=1,r=0; for(i=1;i<=n;i++) { scanf("%d",&x); if(sum<0)sum=x,l=r=i; else sum+=x,r=i; if(sum>ans)ans=sum,ans_l=l,ans_r=r; } printf("%d %d %d\n",ans,ans_l,ans_r); } int main() { freopen("1.in","r",stdin); int t,i; while(scanf("%d",&t)!=EOF) for(i=1;i<=t;i++) { printf("Case %d:\n",i); work(); if(i<t)printf("\n"); } return 0; }
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