HDU 1573~3579 X问题&Hello Kiki[同余方程]

X问题
时限：1000MS
题意很明确，就是让你解一元同余方程组。题目的要求是找出小于等于(N)个数。
利用同余方程的性质，可以找到(X)的最小值(x_0)，同时也知道(Xequiv t(mod\,m)),其中(m)为([m_0,m_1,cdots,m_M]),根据上面的条件可以得到(x_0+m*zleq N)，然后将z解出来就是要的答案。

#include "iostream"
#include "stdio.h"
using namespace std;
typedef long long LL;
LL m[20], r[20];
void ext_gcd(LL a, LL b, LL &d, LL& x, LL& y) {
    LL res;
    if (!b){x = 1; y = 0; d = a;}
    else {
        ext_gcd(b, a%b, d, y, x);
        y -= x*(a/b);
    }
}
LL mod_line(LL a, LL b, LL m, LL& d) {
    LL x, y; ext_gcd(a, m, d, x, y);
    if (b%d) return -1;
    LL e = x*(b/d)%(m/d) + (m/d);
    return e%(m/d);
}
int main(int argc, char const *argv[])
{
    int T;
    LL N, M;
    scanf("%d", &T);
    while (T--) {
        bool flag = false;
        scanf("%lld%lld", &N, &M);
        for (int i = 0; i < M; i++) scanf("%lld", m+i);
        for (int j = 0; j < M; j++) scanf("%lld", r+j);  
        for (int i = 1; i < M; i++) {
            LL d;
            LL x0 = mod_line(m[i-1],r[i] - r[i-1], m[i], d);
            if (flag || x0 == -1) {flag = true;  break;}
            r[i] = x0*m[i-1]+r[i-1];
            m[i] = m[i-1]*(m[i]/d);
            r[i] %= m[i];
        } 
        if (flag) printf("0
");
        else {
            LL t = 0;
            if (r[M-1] <= N) t =(N-r[M-1])/m[M-1] + 1;
            if (t && r[M-1] == 0) t--;   //去掉解是0的情况
            printf("%lld
", t);
        }
        
    }
    return 0;
}

Hello Kiki
时限：1000MS
这道题相比上道题就简单多了，直接计算，然后输出最小的。同样注意要是答案是0的时候的输出。

#include "bits/stdc++.h"
using namespace std;
typedef long long LL;
LL m[20], r[20];
void ext_gcd(LL a, LL b, LL &d, LL& x, LL& y) {
    LL res;
    if (!b){x = 1; y = 0; d = a;}
    else {
        ext_gcd(b, a%b, d, y, x);
        y -= x*(a/b);
    }
}
LL mod_line(LL a, LL b, LL m, LL& d) {
    LL x, y; ext_gcd(a, m, d, x, y);
    if (b%d) return -1;
    LL e = x*(b/d)%(m/d) + (m/d);
    return e%(m/d);
}
int main(int argc, char const *argv[])
{
    int T; LL N;
    int Kcase = 0;
    scanf("%d", &T);
    while (T--) {
        bool flag = false;
        scanf("%lld", &N);
        for (int i = 0; i < N; i++) scanf("%lld", m+i);
        for (int j = 0; j < N; j++) scanf("%lld", r+j);  
        for (int i = 1; i < N; i++) {
            LL d;
            LL x0 = mod_line(m[i-1],r[i] - r[i-1], m[i], d);
            if (flag || x0 == -1) {flag = true;  break;}
            r[i] = x0*m[i-1]+r[i-1];
            m[i] = m[i-1]*(m[i]/d);
            r[i] %= m[i];
        } 
        printf("Case %d: ", ++Kcase);
        if (flag) printf("-1
");
        else {
            if (r[N-1] == 0) r[N-1] = m[N-1];
            printf("%lld
", r[N-1]);
        }
    }
    return 0;
}