hdu6124 Euler theorem 2017多校1005签到题
2017-08-15 19:04
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http://acm.split.hdu.edu.cn/showproblem.php?pid=6124
题意:给定一个数a,求a%b(b为正整数)有多少种可能。
题解:题目说欧拉,差点被骗了,还去把答案和欧拉函数打表出来比对。。。其实就是取b=(a+1)/2,(a+1)/2+1,...,a,可以发现a%b是连续的,并且涵盖了b∈[1,a/2]。最后加上a本身,就是(a+1)/2+1种了。
代码:
#include<bits/stdc++.h>
#define debug cout<<"aaa"<<endl
#define d(a) cout<<a<<endl
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define MIN_INT (-2147483647-1)
#define MAX_INT 2147483647
#define MAX_LL 9223372036854775807i64
#define MIN_LL (-9223372036854775807i64-1)
using namespace std;
const int N = 100000 + 5;
const int mod = 1000000000 + 7;
const double eps = 1e-8;
int main(){
int t,n;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
printf("%d\n",(n+1)/2+1);
}
return 0;
}
题意:给定一个数a,求a%b(b为正整数)有多少种可能。
题解:题目说欧拉,差点被骗了,还去把答案和欧拉函数打表出来比对。。。其实就是取b=(a+1)/2,(a+1)/2+1,...,a,可以发现a%b是连续的,并且涵盖了b∈[1,a/2]。最后加上a本身,就是(a+1)/2+1种了。
代码:
#include<bits/stdc++.h>
#define debug cout<<"aaa"<<endl
#define d(a) cout<<a<<endl
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define MIN_INT (-2147483647-1)
#define MAX_INT 2147483647
#define MAX_LL 9223372036854775807i64
#define MIN_LL (-9223372036854775807i64-1)
using namespace std;
const int N = 100000 + 5;
const int mod = 1000000000 + 7;
const double eps = 1e-8;
int main(){
int t,n;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
printf("%d\n",(n+1)/2+1);
}
return 0;
}
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