HDU 5768:Lucky7(中国剩余定理 + 容斥原理)
http://acm.hdu.edu.cn/showproblem.php?pid=5768
Lucky7
?? once wrote an autobiography, which mentioned something about himself. In his book, it said seven is his favorite number and he thinks that a number can be divisible by seven can bring him good luck. On the other hand, ?? abhors some other prime numbers and thinks a number x divided by pi which is one of these prime numbers with a given remainder ai will bring him bad luck. In this case, many of his lucky numbers are sullied because they can be divisible by 7 and also has a remainder of ai when it is divided by the prime number pi.
Now give you a pair of x and y, and N pairs of ai and pi, please find out how many numbers between x and y can bring ?? good luck.
Each test case starts with three integers three intergers n, x, y(0<=n<=15,0<x<y<1018) on a line where n is the number of pirmes.
Following on n lines each contains two integers pi, ai where pi is the pirme and ?? abhors the numbers have a remainder of ai when they are divided by pi.
It is guranteed that all the pi are distinct and pi!=7.
It is also guaranteed that p1*p2*…*pn<=1018 and 0<ai<pi<=105for every i∈(1…n).
For Case 1: 7,21,42,49,70,84,91 are the seven numbers. For Case2: 7,14,21,28,35,42,49,56,63,70,77,84,91,98 are the fourteen numbers.
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <queue>
#include <cmath>
using namespace std;
typedef long long LL;
#define N 20 LL p[N], a[N];
int bit[N];
int n; LL mul(LL a, LL b, LL m)
{
//快速乘法
LL ans = ;
while(b) {
if(b & ) ans = (ans + a) % m;
a <<= ;
a %= m;
b >>= ;
}
return ans;
} LL exgcd(LL a, LL b, LL &x, LL &y)
{
if(b == ) {
x = ;
y = ;
return a;
}
LL r = exgcd(b, a%b, x, y);
int t = x;
x = y;
y = t - a / b * y;
return r;
} LL CRT(LL x, LL y)
{
//中国剩余定理: 找同时满足多个同余式的解
LL M = , ans = ;
for(int i = ; i <= n; i++) {
if(bit[i]) M *= p[i];
}
for(int i = ; i <= n; i++) {
if(bit[i]) {
LL x, y, Mi;
Mi = M / p[i];
exgcd(Mi, p[i], x, y);
x = (x % p[i] + p[i]) % p[i];
ans = (ans + mul(Mi * a[i] % M, x, M) % M + M) % M;
//ans找出来的是在 M 以内的特解即最小正整数解
}
}
//每过 M 可以有一个解
LL res = (y - ans + M) / M - (x - - ans + M) / M;
return res;
} void solve(LL x, LL y)
{
bit[n] = ;
LL ans = ;
int all = << n;
for(int i = ; i < all; i++) {
int tmp = i, k = ;
for(int j = ; j < n; j++) {
bit[j] = tmp & ;
tmp >>= ;
k += bit[j];
}
k = k & ? - : ;
//k是计算包含多少个同余式
//容斥原理: 奇数减,偶数加,具体可以看《组合数学》P108
//计算出不具有性质(满足任意一个同余式)的数的数量
ans += CRT(x, y) * k;
}
printf("%I64d\n", ans);
} int main()
{
int t;
scanf("%d", &t);
for(int cas = ; cas <= t; cas++) {
LL x, y;
scanf("%d%I64d%I64d", &n, &x, &y);
for(int i = ; i < n; i++)
scanf("%I64d%I64d", &p[i], &a[i]);
p[n] = , a[n] = ;
printf("Case #%d: ", cas);
solve(x, y);
}
return ;
}
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