中国剩余定理的简单应用。。。
#include <iostream>
#include <sstream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <climits>
#define maxn 105
#define eps 1e-6
#define mod 10007
#define INF 99999999
#define lowbit(x) (x&(-x))
//#define lson o<<1, L, mid
//#define rson o<<1 | 1, mid+1, R
//typedef long long LL;
typedef int LL;
using namespace std;
int a[maxn], b[maxn], n, p[1000005];
void extend_gcd(int a, int b, int &d, int &x, int &y)
{
if(b == 0) { d = a, x = 1, y = 0; }
else { extend_gcd(b, a%b, d, y, x), y -= x*(a/b); }
}
void extend_chinese_reminder(int &a1, int &b1)
{
int x, y, g, tmp, i, a2, b2;
for(i = 1; i < n; i++) {
a2 = a[i], b2 = b[i];
extend_gcd(a1, a2, g, x, y);
tmp = a2/g;
x = x*(b2-b1)/g;
x = (x%tmp+tmp)%tmp;
b1 = a1*x+b1;
a1 = (a1*a2)/g;
b1 = (b1%a1+a1)%a1;
}
}
int main(void)
{
int _, m, tmp, i, j, a1, b1, bb;
scanf("%d", &_);
while(_--) {
n = 4, bb = 0;
scanf("%d", &m);
for(i = 0; i < n; i++) scanf("%d", &a[i]);
for(i = 0; i < m; i++) {
scanf("%d", &tmp);
for(j = 0; j < n; j++)
b[3-j] = tmp%100, tmp/=100;
a1 = a[0], b1 = b[0];
extend_chinese_reminder(a1, b1);
for(j = 0; j < n-1; j++)
b[2-j] = b1%100, b1/=100;
for(j = 0; j < n-1; j++)
p[bb++] = b[j];
}
bb--;
for(j = bb; j >= 0; j--) if(p[j]!=27) break;
for(i = 0; i <= j; i++)
if(p[i] == 27) printf(" ");
else printf("%c", p[i]+'A'-1);
printf("\n");
}
return 0;
}
