CometOJ10C 鱼跃龙门

题目链接

problem

实际上就是对于给定的\(n\)求一个最小的\(x\)满足\(\frac{x(x+1)}{2}=kn(k\in N^*)\)

solution

对上面的式子稍微变形可得\(x(x+1)=2kn\)。因为\(x\)\((x+1)\)互质,所以将\(n\)质因数分解后,同种质因子肯定都位于\(x\)\((x+1)\)中。\(10^{12}\)以内的整数质因数分解后种类不超过\(13\),所以可以暴力枚举每种质因子属于\(x\)还是\(x+1\)

然后分别得到\(a\)\(b\)。下面要使得\(bx=ay+1\)。扩展欧几里得求解即可。

PS

本题时限\(0.5s\),每次询问都\(\sqrt{n}\)质因数分解是会\(TLE\)的。所以先预处理质数。然后进行质因数分解。

code

//@Author: wxyww
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
#include<cmath>
#include<map>
#include<string>
using namespace std;
typedef long long ll;
const int N = 5000010;
ll read() {
    ll x = 0,f = 1; char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0',c = getchar();}
    return x * f;
}
ll dis[N];
int prmjs,vis[N];
ll n;
ll js,cnt[15];
void fj(ll x) {
    for(int i = 1;dis[i] * dis[i] <= x;++i) {
        if(x % dis[i] == 0) {
            cnt[++js] *= dis[i];
            x /= dis[i];
        }
        while(x % dis[i] == 0) x /= dis[i],cnt[js] *= dis[i];
    }
    if(x != 1) cnt[++js] = x;
}
ll ans;
ll exgcd(ll a,ll b,ll &x,ll &y) {
   if(b == 0) {
      x = 1,y = 0;return a;
    }
   ll tmp = exgcd(b,a % b,x,y);
   ll t = x;
   x = y; y = t - a / b * y;
   return tmp;
}

int main() {
    int T = read();
    for(int i = 2;i <= 2000000;++i) {
        if(!vis[i]) dis[++prmjs] = i;
        for(int j = 1;j <= prmjs && 1ll * dis[j] * i <= 1000000;++j) {
            vis[dis[j] * i] = 1;
            if(i % dis[j] == 0) break;
        }
    }
    while(T--) {
        n = read() * 2;
        js = 0;
        for(int i = 1;i <= 14;++i) cnt[i] = 1;
        fj(n);
        ans = n * 2;
        ll m = 1 << js;
        for(int i = 0;i < m;++i) {
            ll now = 1;
            for(int j = 0;j < js;++j) if(i >> j & 1) now *= cnt[j + 1];
            ll x,y;
            exgcd(now,n / now,x,y);
            y = y % now;
            if(y >= 0) y -= now;
            ans = min(ans,n / now * -y);
        }
        printf("%lld\n",ans);
    }
    return 0;
}
posted @ 2019-09-07 16:46 wxyww 阅读(...) 评论(...) 编辑 收藏