Lucky7
Problem Description
When ?? was born, seven crows flew in and stopped beside him. In its childhood, ?? had been unfortunately fall into the sea. While it was dying, seven dolphins arched its body and sent it back to the shore. It is said that ?? used to surrounded by 7 candles when he faced a extremely difficult problem, and always solve it in seven minutes. ?? 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.
Input
On the first line there is an integer T(T≤20) representing the number of test cases. 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).
Output
For each test case, first output "Case #x: ",x=1,2,3...., then output the correct answer on a line.
Sample Input
2
2 1 100
3 2
5 3
0 1 100
Sample Output
Case #1: 7
Case #2: 14
Hint
题意：找出[l, r]里面可以被 7 整除的并且不满足任意一个同余式的数的个数。
1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 using namespace std; 8 typedef long long LL; 9 #define N 2010 11 LL p[N], a[N];12 int bit[N];13 int n;14 15 LL mul(LL a, LL b, LL m)16 {17 //快速乘法18 LL ans = 0;19 while(b) {20 if(b & 1) ans = (ans + a) % m;21 a <<= 1;22 a %= m;23 b >>= 1;24 }25 return ans;26 }27 28 LL exgcd(LL a, LL b, LL &x, LL &y)29 {30 if(b == 0) {31 x = 1;32 y = 0;33 return a;34 }35 LL r = exgcd(b, a%b, x, y);36 int t = x;37 x = y;38 y = t - a / b * y;39 return r;40 }41 42 LL CRT(LL x, LL y)43 {44 //中国剩余定理: 找同时满足多个同余式的解45 LL M = 1, ans = 0;46 for(int i = 0; i <= n; i++) {47 if(bit[i]) M *= p[i];48 }49 for(int i = 0; i <= n; i++) {50 if(bit[i]) {51 LL x, y, Mi;52 Mi = M / p[i];53 exgcd(Mi, p[i], x, y);54 x = (x % p[i] + p[i]) % p[i];55 ans = (ans + mul(Mi * a[i] % M, x, M) % M + M) % M;56 //ans找出来的是在 M 以内的特解即最小正整数解57 }58 }59 //每过 M 可以有一个解60 LL res = (y - ans + M) / M - (x - 1 - ans + M) / M;61 return res;62 }63 64 void solve(LL x, LL y)65 {66 bit[n] = 1;67 LL ans = 0;68 int all = 1 << n;69 for(int i = 0; i < all; i++) {70 int tmp = i, k = 0;71 for(int j = 0; j < n; j++) {72 bit[j] = tmp & 1;73 tmp >>= 1;74 k += bit[j];75 }76 k = k & 1 ? -1 : 1;77 //k是计算包含多少个同余式78 //容斥原理: 奇数减,偶数加,具体可以看《组合数学》P10879 //计算出不具有性质（满足任意一个同余式）的数的数量80 ans += CRT(x, y) * k;81 }82 printf("%I64d\n", ans);83 }84 85 int main()86 {87 int t;88 scanf("%d", &t);89 for(int cas = 1; cas <= t; cas++) {90 LL x, y;91 scanf("%d%I64d%I64d", &n, &x, &y);92 for(int i = 0; i < n; i++)93 scanf("%I64d%I64d", &p[i], &a[i]);94 p[n] = 7, a[n] = 0;95 printf("Case #%d: ", cas);96 solve(x, y);97 }98 return 0;99 }