记录编号 |
237347 |
评测结果 |
AAAAAAAAAA |
题目名称 |
无关的数 |
最终得分 |
100 |
用户昵称 |
Fmuckss |
是否通过 |
通过 |
代码语言 |
C++ |
运行时间 |
0.154 s |
提交时间 |
2016-03-16 22:41:16 |
内存使用 |
0.28 MiB |
显示代码纯文本
#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
typedef long long LL;
LL n, m;
LL gcd(LL a, LL b) {
return b == 0 ? a : gcd(b, a%b);
}
vector<LL> sol;
LL up, down;
void handle1(LL x) {
LL tmp = gcd(x, m);
while(tmp != 1 && x != 1) {//tmp == 1 意味着两个数没有最大公约数
up *= tmp;
x /= tmp;
tmp = gcd(x, m);
}
}
void handle2(LL x) {
LL tmp = gcd(x, m);
while(tmp != 1 && x != 1) {
down *= tmp;
x /= tmp;
tmp = gcd(x, m);
}
}
int main() {
freopen("irre.in", "r", stdin);
freopen("irre.out", "w", stdout);
scanf("%lld %lld", &n, &m);//要求第n行的数,其实是求c(n-1, i)
//即对于n的第2~n/2位处理,如9,2=> c(8,1) 9,3=>c(8,2)
up = 1; down = 1;
LL tmp;
for(LL i = 2; i < n; i++) {//从第2到第n-1 n,i c(n-1, i-1)
handle1(n-i+1);
handle2(i-1);
tmp = gcd(up, down);
while(tmp != 1 && up != 1 && down != 1) {
up /= tmp;
down /= tmp;
tmp = gcd(up, down);
}
if(gcd(up, m) == m) {
sol.push_back(i);
}
}
printf("%d\n",sol.size());
for(LL i = 0 ; i < sol.size(); i++) {
printf("%lld ",sol[i]);
}
return 0;
}