标签:poj
转载请注明出处:http://blog.csdn.net/u012860063?viewmode=contents
题目链接:http://poj.org/problem?id=3077
Description
Input
Output
Sample Input
9 15 14 4 5 99 12345678 44444445 1445 446
Sample Output
20 10 4 5 100 10000000 50000000 2000 500
代码一如下:
#include <iostream>
#include <cstring>
using namespace std;
int main()
{
int t, n, k, count;
char s[17];
int i, j;
while(cin >>t)
{
while(t--)
{
count = 0;
int p = 0, l = 0;;
memset(s,0,sizeof(s));
cin>>s;
int len = strlen(s);
if(len == 1)
{
cout<<s[0]<<endl;
continue;
}
for(i = len-1; i > 0; i--)
{
if(s[i]-'0'+p > 4)
{
p = 1;
count++;
}
else
{
p = 0;
count++;
}
}
if(s[0]-'0' + p > 9)
{
cout<<10;
}
else
{
cout<<s[0]-'0'+p;
}
for(i = 0; i < count; i++)
{
cout<<'0';
}
cout<<endl;
}
}
return 0;
}#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
using namespace std;
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
int n, count = 0;
scanf("%d", &n);
double x = n;
while (x >= 10)
{
x /= 10;
x = (int)(x + 0.5);
count++;
}
n = (int)x;
for (int i = 0; i < count; i++)
n *= 10;
printf("%d\n", n);
}
return 0;
}poj3077Rounders(模拟),布布扣,bubuko.com
标签:poj
原文地址:http://blog.csdn.net/u012860063/article/details/37700403