数论 2016.4.9

1、FZU 1402 猪的安家

典型中国剩余定理

#include <iostream>#include <cstdio>using namespace std;typedef long long LL;const int maxn = 1000 + 5;int a[maxn], b[maxn];LL x, y;LL Ext_Gcd(LL a, LL b, LL &x, LL &y);int main(){//    freopen("in.txt", "r", stdin);    int n;    while (cin>>n) {        LL mul = 1;        for (int i=0; i<n; ++i) {            cin>>a[i]>>b[i];            mul *= a[i];        }        LL ans = 0;        for (int i=0; i<n; ++i) {            LL t = mul/a[i];            Ext_Gcd(t, a[i], x, y);            x = (x%a[i] + a[i]) % a[i];            ans = (ans + t * x * b[i]) % mul;        }        cout<<ans<<endl;    }    return 0;}LL Ext_Gcd(LL a, LL b, LL &x, LL &y){    if (b == 0) {        x = 1;        y = 0;        return a;    }    int d = Ext_Gcd(b, a%b, x, y);    int t = x;    x = y;    y = t - a/b*y;    return d;}

2、POJ 1995 Raising Modulo Numbers

最简单的快速幂

#include <iostream>#include <cstdio>#include <cstring>using namespace std;typedef long long LL;LL pow_mod(LL x, LL n, LL mod_val);int main(){//    freopen("in.txt", "r", stdin);    int Z;    scanf("%d", &Z);    while (Z--) {        int ans = 0;        int M;        scanf("%d", &M);        int H;        scanf("%d", &H);        int A, B;        for (int i=0; i<H; ++i) {            scanf("%d%d", &A, &B);            ans = (ans + pow_mod(A%M, B, M)) % M;        }        printf("%d\n", ans);    }    return 0;}LL pow_mod(LL x, LL n, LL mod_val){    LL ret = 1;    LL t = x % mod_val;    while (n) {        if (n&1) {            ret = (ret * t) % mod_val;        }        t = (t * t) % mod_val;        n >>= 1;    }    return ret;}

3、UVa 408 Uniform Generator

gcd

#include <iostream>#include <cstdio>#include <cstring>#include <iomanip>using namespace std;int Gcd(int a, int b);int main(){//    freopen("in.txt", "r", stdin);    int a, b;    while (cin>>a>>b) {        if (Gcd(a, b) == 1) {            cout<<setw(10)<<setfill(' ')<<a;            cout<<setw(10)<<setfill(' ')<<b;            cout<<"    Good Choice"<<endl;        } else {            cout<<setw(10)<<setfill(' ')<<a;            cout<<setw(10)<<setfill(' ')<<b;            cout<<"    Bad Choice"<<endl;        }        cout<<endl;    }    return 0;}int Gcd(int a, int b){    return ((a%b == 0) ? b : Gcd(b, a%b));}

4、HDU 2098 分拆素数和

素数筛，可暴力

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 10000 + 10;bool IsPrime[maxn];void Init(void);int main(){//    freopen("in.txt", "r", stdin);    int N;    Init();    while (cin>>N && N!=0) {        int Count = 0;        for (int i=2; i<N/2; ++i) {            if (IsPrime[i] && IsPrime[N-i]) {                ++Count;            }        }        cout<<Count<<endl;    }    return 0;}void Init(void){    memset(IsPrime, true, sizeof(IsPrime));    for (int i=2; i<=maxn; ++i) {        if (IsPrime[i]) {            int j = 2;            while (i*j <= maxn) {                IsPrime[i*j] = false;                ++j;            }        }    }}

5、HDU 1061 Rightmost Digit

稳定的快速幂模版

#include <iostream>using namespace std;typedef long long LL;LL pow_mod(LL x, LL n, LL mod_val);LL mult_mod(LL x, LL y, LL mod_val);int main(){    int T;    int N;    while (cin>>T) {        while (T--) {            cin>>N;            cout<<pow_mod(N, N, 10)<<endl;        }    }    return 0;}LL mult_mod(LL a, LL b, LL mod_val){    a %= mod_val;    b %= mod_val;    LL ret = 0;    LL t = a;    while (b) {        if (b&1) {            ret += t;            if (ret > mod_val) {                ret -= mod_val;//直接取模慢很多            }        }        t <<= 1;        if (t > mod_val) {            t -= mod_val;        }        b >>= 1;    }    return ret;}// 计算 ret = (x^n)%mod_valLL pow_mod(LL x, LL n, LL mod_val){    LL ret = 1;    LL t = x%mod_val;    while (n)    {        if (n&1) {            ret = mult_mod(ret, t, mod_val);        }        t = mult_mod(t, t, mod_val);        n >>= 1;    }    return ret;}