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Description

輸入格式

要求由鍵盤輸入A,B,C,D,E,F,G,H,a,b,c,d,e,f,g,h十六個數，分別代表每A人一列餘a、每B人一列餘b、每C人一列餘c、每D人一列餘D、每E人一列餘e、每F人一列餘f、每G人一列餘g、每H人一列餘h，其中A,B,C,D,E,F,G,H為互不相等的質數

輸出格式

輸出總兵士數，要求輸出滿足條件的最小的一個，但要滿足8種排法的每一種排法至少可排一列。（保證給的資料，有結果且計算的結果不會超過2的63次方）

輸入樣例

2 3 5 7 11 13 17 19
1 1 1 1 1 1 1 1

輸出樣例

9699691

暑假看了點數論，在網上看了些中國剩餘定理，回頭寫這道數學題/呲牙

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <string>
#include <stack>
#include <set>
#include <map>
#include <iostream>
#include <bitset>
#include <algorithm>
using namespace std;

#define MP make_pair
#define PB push_back
#define mst(a,b) memset((a),(b),sizeof(a))
#define TEST cout<<"*************************"<<endl

#define rep(s,n,up) for(int i = (s); i < (n); i+=(up))
#define per(n,e,down) for(int i = (n); i >= (e); i-=(down))
#define rep1(s,n,up) for(int j = (s); j < (n); j+=(up))
#define per1(n,e,down) for(int j = (n); j >= (e); j-=(down))

typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> Pii;
typedef vector<int> Vi;
typedef vector<Pii> Vii;
const int inf = 0x3f3f3f3f;
const LL INF = (1uLL << 63) - 1;
const double Pi = acos(-1.0);
const int maxn = (1 << 16) + 7;
const uLL Hashmod = 29050993;
const double esp=1e-6;


//#define local


int main() {
#ifdef local
    freopen("input.txt", "r", stdin);
    //freopen("output.txt","w",stdout);
#endif
    //ios::sync_with_stdio(0);
    //cin.tie();
    LL divisor[8],remainder[8],tot_m=1,sum=0;
    rep(0,8,1)scanf("%lld",&divisor[i]);
    rep(0,8,1)scanf("%lld",&remainder[i]);
    LL min_sum=divisor[0]+remainder[0];
    rep(0,8,1){
        tot_m*=divisor[i];
        LL s=divisor[i]+remainder[i];
        if(s>min_sum)
            min_sum=s;
    }
    rep(0,8,1){
        LL m=1,temp_m=1;
        rep1(0,8,1){
            if(j!=i)
                m*=divisor[j];
        }
        temp_m=m;
        while(m%divisor[i]!=1)
            m+=temp_m;
        sum+=m*remainder[i];
    }
    sum%=tot_m;
    while(sum<min_sum)sum+=tot_m;
    printf("%lld\n",sum);
    return 0;
}