hdu 1003 Max Sum
来源:互联网 发布:大芒果数据库 编辑:程序博客网 时间:2024/06/11 23:55
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 143429 Accepted Submission(s): 33431
Problem DescriptionGiven 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.
InputThe 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).
OutputFor 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 Input25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5
Sample OutputCase 1:14 1 4Case 2:7 1 6
#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;int main (){ int T,n,a[100105]; int i,j,sum,max,s,e,t; cin>>T; for(i=1;i<=T;i++) { cin>>n; for(j=0;j<n;j++) cin>>a[j]; sum=a[0]; max=a[0]; t=0; s=0; e=0; for(j=1;j<n;j++) { if(sum<0) { sum=0; t=j; } sum+=a[j]; if(sum>max) { max=sum; e=j; s=t; } } cout<<"Case"<<" "<<i<<":"<<endl; cout<<max<<" "<<s+1<<" "<<e+1<<endl; if(i<T) cout<<endl; } return 0;}
25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5
Case 1:14 1 4Case 2:7 1 6
#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;int main (){ int T,n,a[100105]; int i,j,sum,max,s,e,t; cin>>T; for(i=1;i<=T;i++) { cin>>n; for(j=0;j<n;j++) cin>>a[j]; sum=a[0]; max=a[0]; t=0; s=0; e=0; for(j=1;j<n;j++) { if(sum<0) { sum=0; t=j; } sum+=a[j]; if(sum>max) { max=sum; e=j; s=t; } } cout<<"Case"<<" "<<i<<":"<<endl; cout<<max<<" "<<s+1<<" "<<e+1<<endl; if(i<T) cout<<endl; } return 0;}
0 0
- HDU 1003 Max Sum
- hdu 1003 Max Sum
- hdu 1003 Max Sum
- HDU-1003 max sum
- HDU 1003 - Max Sum
- HDU 1003 Max Sum
- hdu 1003 Max Sum
- HDU 1003 Max Sum
- HDU 1003 Max Sum
- hdu 1003 max sum
- HDU 1003 Max Sum
- hdu 1003 Max Sum
- HDU 1003 Max Sum
- hdu 1003 Max Sum
- HDU 1003 Max Sum
- Hdu 1003 - Max Sum
- HDU-1003-Max Sum
- hdu - 1003 - Max Sum
- 终于把cocos3.2 对应CocosStudio的按钮回调写出来了。
- 男人患罕见“臭鱼症” 满身腥臭体味像死鱼
- 使用SQL语句创建和删除约束
- 李瑞英证实与张宏平易近辞别《往事联播》
- n个数的最小公倍数
- hdu 1003 Max Sum
- hdu 4512 DP
- 王菲微博“逗贫”语录暴光
- 求数组中出现多于一半的数字
- 【POJ】3041 Asteroids 二分匹配
- hdu 1115 Lifting the Stone(计算几何(重心))
- 更改 centos yum 源
- POJ1128 Frame Stacking 【拓扑排序】+【深搜】
- 精通css(9)bug和修复bug