hdu1573 X问题

来源：互联网发布：哪里可以买淘宝号编辑：程序博客网时间：2024/05/21 09:07

原题：

http://acm.hdu.edu.cn/showproblem.php?pid=1573

这个题目真是坑爹啊- =首先说M个正整数，结果测试数据里面就有0 - =莫非0成了正整数？
扩展中国剩余定理解这道题目很多模板也不对- =造呢。。也不知道到底可不可以从0开始- =坑爹的题目。。

如果时间急别在这里干着急了- =坑死人

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define lln long long intlln a[11];lln b[11];lln gcd( lln a, lln b){return b == 0 ? a : gcd(b, a % b);}void ace(){int n, m;int t;int i, j;int lcm;bool flag;int num;cin >> t;while (t--){memset(a, 0, sizeof(a));memset(b, 0, sizeof(b));cin >> n >> m;lcm = 1;flag = false;num = 0;for (i = 0; i < m; i++){cin >> a[i];lcm = lcm * a[i] / gcd(lcm, a[i]);}for (i = 0; i < m; i++)cin >> b[i];for (i = 1; i <= lcm && i <= n; i++) //find the minium num 坑爹的地方- - i不能从0开始{for (j = 0; j < m; j++)if (i % a[j] != b[j])break;if (j == m){flag = true;break;}}if (flag){int temp = n % lcm;if (temp >= i)i = n / lcm + 1;elsei = n / lcm;printf("%d\n", i);//while (i <= n)//{//i += lcm;//num++;//}//cout << num << endl;}elseprintf("0\n");}}int main(){ace();return 0;}

/**中国剩余定理（不互质）模板*/#include <iostream>#include <cstdio>#include <cstring>using namespace std;typedef __int64 int64;int64 Mod;int64 gcd(int64 a, int64 b){    if(b==0)        return a;    return gcd(b,a%b);}int64 Extend_Euclid(int64 a, int64 b, int64&x, int64& y){    if(b==0)    {        x=1,y=0;        return a;    }    int64 d = Extend_Euclid(b,a%b,x,y);    int64 t = x;    x = y;    y = t - a/b*y;    return d;}//a在模n乘法下的逆元，没有则返回-1int64 inv(int64 a, int64 n){    int64 x,y;    int64 t = Extend_Euclid(a,n,x,y);    if(t != 1)        return -1;    return (x%n+n)%n;}//将两个方程合并为一个bool merge(int64 a1, int64 n1, int64 a2, int64 n2, int64& a3, int64& n3){    int64 d = gcd(n1,n2);    int64 c = a2-a1;    if(c%d)        return false;    c = (c%n2+n2)%n2;    c /= d;    n1 /= d;    n2 /= d;    c *= inv(n1,n2);    c %= n2;    c *= n1*d;    c += a1;    n3 = n1*n2*d;    a3 = (c%n3+n3)%n3;    return true;}//求模线性方程组x=ai(mod ni),ni可以不互质int64 China_Reminder2(int len, int64* a, int64* n){    int64 a1=a[0],n1=n[0];    int64 a2,n2;    for(int i = 1; i < len; i++)    {        int64 aa,nn;        a2 = a[i],n2=n[i];        if(!merge(a1,n1,a2,n2,aa,nn))            return -1;        a1 = aa;        n1 = nn;    }    Mod = n1;    return (a1%n1+n1)%n1;}int64 a[1000],b[1000];int main(){    int i;    int k;    while(scanf("%d",&k)!=EOF)    {        for(i = 0; i < k; i++)            scanf("%I64d %I64d",&a[i],&b[i]);        printf("%I64d\n",China_Reminder2(k,b,a));    }    return 0;}

0 0