A. The number of positions
2013-03-10 20:45
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time limit per test
0.5 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Petr stands in line of n people, but he doesn't know exactly which position he occupies. He can say that there are no less thana people
standing in front of him and no more than b people standing behind him. Find the number of different positions Petr can occupy.
Input
The only line contains three integers n, a and b (0 ≤ a, b < n ≤ 100).
Output
Print the single number — the number of the sought positions.
Sample test(s)
input
output
input
output
Note
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5.
解题说明:这道题就是考虑前a个不能站人,同时要考虑后面的人不能超过b个,分情况判断即可,当总人数过多或a太小时,以b为准,否则就以a为准
#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{
int n,a,b;
int temp1;
int answer;
scanf("%d %d %d",&n,&a,&b);
temp1=n-a;
if(temp1-1>b)
{
answer=b+1;
}
else
{
answer=temp1;
}
printf("%d\n",answer);
return 0;
}
0.5 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Petr stands in line of n people, but he doesn't know exactly which position he occupies. He can say that there are no less thana people
standing in front of him and no more than b people standing behind him. Find the number of different positions Petr can occupy.
Input
The only line contains three integers n, a and b (0 ≤ a, b < n ≤ 100).
Output
Print the single number — the number of the sought positions.
Sample test(s)
input
3 1 1
output
2
input
5 2 3
output
3
Note
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5.
解题说明:这道题就是考虑前a个不能站人,同时要考虑后面的人不能超过b个,分情况判断即可,当总人数过多或a太小时,以b为准,否则就以a为准
#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{
int n,a,b;
int temp1;
int answer;
scanf("%d %d %d",&n,&a,&b);
temp1=n-a;
if(temp1-1>b)
{
answer=b+1;
}
else
{
answer=temp1;
}
printf("%d\n",answer);
return 0;
}
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