ZOJ - 2781 Rounders
2015-07-27 10:10
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Rounders
Submit Status
Description
Introduction
For a given number, if greater than ten, round it to the nearest ten, then (if that result is greater than 100) take the result and round it to the nearest hundred, then (if that result is greater than 1000) take that number and round it to the nearest thousand,
and so on ...
Input
Input to this problem will begin with a line containing a single integer n indicating the number of integers to round. The next n lines each contain a single integer x (0 <= x <= 99999999).
Output
For each integer in the input, display the rounded integer on its own line.
Note: Round up on fives.
Sample Input
Sample Output
Source
South Central USA Regional Programming Contest 2006
分析:
水题。
ac代码:
#include <iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
int n,i,x,x2;
int c;
scanf("%d",&n);
while(n--)
{
scanf("%d",&x);
c=0;
x2=x;
for(;x2;x2/=10)
{
c++;
}
//printf("c=%d\n",c);
for(i=1;i<c;i++)
{
if(x%(int)(pow(10,i))>=(int)5*(pow(10,i-1)))
{
x=x-x%(int)(pow(10,i))+(int)(pow(10,i));
//printf("xx=%d",x);
}
else
{
x=x-x%(int)(pow(10,i));
//printf("xx=%d",x);
}
}
printf("%d\n",x);
}
return 0;
}
Time Limit: 2000MS | Memory Limit: 65536KB | 64bit IO Format: %lld & %llu |
Description
Introduction
For a given number, if greater than ten, round it to the nearest ten, then (if that result is greater than 100) take the result and round it to the nearest hundred, then (if that result is greater than 1000) take that number and round it to the nearest thousand,
and so on ...
Input
Input to this problem will begin with a line containing a single integer n indicating the number of integers to round. The next n lines each contain a single integer x (0 <= x <= 99999999).
Output
For each integer in the input, display the rounded integer on its own line.
Note: Round up on fives.
Sample Input
9 15 14 4 5 99 12345678 44444445 1445 446
Sample Output
20 10 4 5 100 10000000 50000000 2000 500
Source
South Central USA Regional Programming Contest 2006
分析:
水题。
ac代码:
#include <iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
int n,i,x,x2;
int c;
scanf("%d",&n);
while(n--)
{
scanf("%d",&x);
c=0;
x2=x;
for(;x2;x2/=10)
{
c++;
}
//printf("c=%d\n",c);
for(i=1;i<c;i++)
{
if(x%(int)(pow(10,i))>=(int)5*(pow(10,i-1)))
{
x=x-x%(int)(pow(10,i))+(int)(pow(10,i));
//printf("xx=%d",x);
}
else
{
x=x-x%(int)(pow(10,i));
//printf("xx=%d",x);
}
}
printf("%d\n",x);
}
return 0;
}
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