扩展欧几里德定理

来源：互联网发布：少儿编程培训班编辑：程序博客网时间：2024/05/19 10:34

hdu 4596 Yet another end of the world

第一次接触这类题目，主要借鉴该博文 http://blog.csdn.net/pi9nc/article/details/22911885

#include<cstdio>#include<algorithm>#include<vector>#include<cmath>#include<set>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;const int MAXN = 1005;int x[MAXN], y[MAXN], z[MAXN], n;int gcd(int a, int b){return b?gcd(b,a%b):a;}int judge(int a, int l, int r){if (l%a == 0 || r%a == 0) return 1;if (l < 0 && r > 0) return 1;if (r/a - l/a >= 1) return 1;return 0;}int solve(){for (int i = 0; i< n; ++i){for (int j = i+1; j< n; ++j){int g = gcd(x[i], x[j]);if (judge(g,y[i]-z[j],z[i]-y[j])) return 1;}}return 0;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);#endifwhile (scanf("%d", &n) != EOF){for (int i = 0; i< n; ++i) scanf("%d%d%d", x+i, y+i, z+i);if (solve()) puts("Cannot Take off");else puts("Can Take off");}return 0;}

hdu 1576 A/B

上题用扩展欧几里德定理中判别一阶不定方程是否有解，该题则在原来的基础上求方程的解；
p * a + q * b = gcd(a, b) ......(0)
A = 9973*c + n
A = B*k
n = B * k - 9973*c ......(1)
∵ gcd(B, 9973) == 1
∴ 方程(1)恒有解
先找出一组解 p0, q0（p0为0附近最小解）
则推出其他的解为：
p = p0 + b/gcd(a,b)*t
q = q0 + a/gcd(a,b)*t
在找出第一组解 k0，c0后(k0可能小于0)，ki = k0+9973*t;
∵ 最后结果为 k % 9973
∴ return (k0%mod + mod)%mod*n%mod
#include<cstdio>#include<algorithm>#include<vector>#include<cmath>#include<set>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;typedef __int64 ll;const ll mod = 9973;template<class T>T exGCD(T a, T b, T &x, T &y){if (b == 0){x = 1;y = 0;return a;}T r = exGCD(b,a%b,x,y);T t = x;x = y;y = t-a/b*y;return r;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);#endifint t;scanf("%d", &t);while (t--){ll n, B, k, c;scanf("%I64d%I64d", &n, &B);exGCD(B, mod, k, c);k = (k%mod + mod)%mod;printf("%I64d\n", n*k%mod);}return 0;}

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