周赛二 简单题目汇总
1、http://acm.hdu.edu.cn/showproblem.php?pid=2113
Secret Number
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7369 Accepted Submission(s): 3124
Problem Description
有一天, KIKI 收到一张奇怪的信, 信上要KIKI 计算出给定数各个位上数字为偶数的和.
eg. 5548
结果为12 , 等于 4 + 8
KIKI 很苦恼. 想请你帮忙解决这个问题.
eg. 5548
结果为12 , 等于 4 + 8
KIKI 很苦恼. 想请你帮忙解决这个问题.
Input
输入数据有多组,每组占一行,只有一个数字,保证数字在INT范围内.
Output
对于每组输入数据,输出一行,每两组数据之间有一个空行.
Sample Input
415326 3262
Sample Output
12 10
Author
kiki
AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int n,flag=0;
while(scanf("%d",&n)!=EOF)
{
int k,sum=0;
while(n)
{
int k=n%10;
if(k%2==0)
sum+=k;
n/=10;
}
if(flag==0)
{
printf("%d\n",sum);
flag=1;
}
else
{
printf("\n");
printf("%d\n",sum);
}
}
return 0;
}
2、http://acm.hdu.edu.cn/showproblem.php?pid=2115
题目：
I Love This Game
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5247 Accepted Submission(s): 1810
Problem Description
Do you like playing basketball ? If you are , you may know the NBA Skills Challenge . It is the content of the basketball skills . It include several parts , such as passing , shooting , and so on. After completion of the content , the player who takes the shortest time will be the winner . Now give you their names and the time of finishing the competition , your task is to give out the rank of them ; please output their name and the rank, if they have the same time , the rank of them will be the same ,but you should output their names in lexicographic order.You may assume the names of the players are unique.
Is it a very simple problem for you? Please accept it in ten minutes.
Is it a very simple problem for you? Please accept it in ten minutes.
Input
This problem contains multiple test cases! Ease test case contain a n(1<=n<=10) shows the number of players,then n lines will be given. Each line will contain the name of player and the time(mm:ss) of their finish.The end of the input will be indicated by an integer value of zero.
Output
The output format is shown as sample below.
Please output the rank of all players, the output format is shown as sample below;
Output a blank line between two cases.
Please output the rank of all players, the output format is shown as sample below;
Output a blank line between two cases.
Sample Input
10 Iverson 17:19 Bryant 07:03 Nash 09:33 Wade 07:03 Davies 11:13 Carter 14:28 Jordan 29:34 James 20:48 Parker 24:49 Kidd 26:46 0
Sample Output
Case #1 Bryant 1 Wade 1 Nash 3 Davies 4 Carter 5 Iverson 6 James 7 Parker 8 Kidd 9 Jordan 10
Author
為傑沉倫
Source
HDU 2007-10 Programming Contest_WarmUp
Recommend
威士忌
Ac代码：
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 15
struct node
{
string name;
int h;
int m;
int r;
}a[N];
int cmp(node b,node c)
{
if(b.h==c.h && b.m==c.m)
return b.name<c.name;
if(b.h==c.h)
return b.m<c.m;
return b.h<c.h;
}
int main()
{
int n,cas=0;
char ch;
while(scanf("%d",&n)!=EOF)
{
if(n==0)
break;
cas++;
if(cas!=1)
{
printf("\n");
}
for(int i=1;i<=n;i++)
{
cin>>a[i].name>>a[i].h>>ch>>a[i].m;
//scanf("%s%d:%d",a[i].name,&a[i].h,&a[i].m);
}
sort(a+1,a+n+1,cmp);
printf("Case #%d\n",cas);
//printf("%s 1\n",a[1].name);
cout<<a[1].name<<" "<<1<<endl;
// a[1].r=1;
int m=1;
for(int i=2;i<=n;i++)
{
if(a[i].h==a[i-1].h && a[i].m==a[i-1].m)
{
//printf("%s %d\n",a[i].name,a[i-1].r);
cout<<a[i].name<<" "<<m<<endl;
//a[i].r=a[i-1].r;
}
else
{
//printf("s %d\n",a[i].name,i);
cout<<a[i].name<<" "<<i<<endl;
m=i;
}
}
}
return 0;
}
/*
5
ab 00:00
a 01:00
b 01:00
abc 01:10
aa 01:10
*/
3、http://acm.hdu.edu.cn/showproblem.php?pid=2117
Just a Numble
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2072 Accepted Submission(s): 968
Problem Description
Now give you two integers n m, you just tell me the m-th number after radix point in 1/n,for example n=4,the first numble after point is 2,the second is 5,and all 0 followed
Input
Each line of input will contain a pair of integers for n and m(1<=n<=10^7,1<=m<=10^5)
Output
For each line of input, your program should print a numble on a line,according to the above rules
Sample Input
4 2 5 7 123 123
Sample Output
5 0 8
Author
YBB
Source
HDU 2007-10 Programming Contest_WarmUp
Recommend
威士忌
Ac代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
int k=1,j=0,sum=1;
for(int i=1;i<=m+1;i++)
{
k=sum/n;
j=sum%n;
sum=j*10;
}
printf("%d\n",k);
}
return 0;
}