Lucky number__zju3233 解题报告
Lucky Number
Time Limit: 5 Seconds Memory Limit: 32768 KB
Watashi loves M mm very much. One day, M mm gives Watashi a chance to choose a number between low and high, and if the choosen number is lucky, M mm will marry him.
M mm has 2 sequences, the first one is BLN (Basic Lucky Numbers), and the second one is BUN (Basic Unlucky Numbers). She says that a number is lucky if it's divisible by at least one number from BLN and not divisible by at least one number from BUN.
Obviously, Watashi doesn't know the numbers in these 2 sequences, and he asks M mm that how many lucky number are there in [low, high]?
Please help M mm calculate it, meanwhile, tell Watashi what is the probability that M mm marries him.
Input
The first line of each test case contains the numbers NBLN (1 <= NBLN <= 15), NBUN (1 <= NBUN <= 500), low, high (1 <= low <= high <= 1018).
The second and third line contain NBLN and NBUN integers, respectively. Each integer in sequences BLN and BUN is from interval [1, 32767].
The last test case is followed by four zero.
The input will contain no more than 50 test cases.
Output
For each test case output one number, the number of lucky number between low and high.
Sample Input
2 1 70 81 2 3 5 0 0 0 0
Sample Output
5
Hint
The lucky numbers in the sample are 72, 74, 76, 78, 81.
解题思路：
命题A：至少能被BLN中的一个数整除
命题B：至少不能被BUN中的一个数整除
所以A&&B=A-A&&(~B)；
题目就是求[low,high]中符合A&&B的数的个数，用容斥原理求出符合A的个数并减去符合A&&(~B)的个数；
求出BUN中所有数的最小公倍数p，
递归从BLN中取出k个数，求他们的最小公倍数g,
求出p和g的最小公倍数s；
当k为奇数：ans+=high/g-high/s-((low-1)/g-(low-1)/s);
当k为偶数：ans-=high/g-high/s-((low-1)/g-(low-1)/s);（容斥原理）
ans就是最后结果；
时间复杂度：2^15
浙大oj 840ms
空间复杂度 180kB
代码：
#include<stdio.h> #define BG 1000000000000000000LL long long ans,l,h,p,n; long long a[18],b[505]; long long gcd(long long a,long long b) { if (b==0) return a; else return gcd(b,a%b); } long long calc(long long g,long long k,long long t) { long long gg;long long s; if (t==n+1) { if (k!=0) { s=gcd(p,g); if (p/s>BG/g) s=h+1; else s=p/s*g; if (k%2==0) ans-=h/g-h/(s)-((l-1)/g-(l-1)/(s)); else ans+=h/g-h/(s)-((l-1)/g-(l-1)/(s)); } return 0; } gg=gcd(g,a[t]); if (g/gg>BG/a[t]) gg=h+1; else gg=g/gg*a[t]; calc(gg,k+1,t+1); calc(g,k,t+1); } int main() { long long m,i,j,g,k; while(scanf("%lld%lld%lld%lld",&n,&m,&l,&h)!=EOF) { if ((n==0)&&(m==0)&&(l==0)&&(h==0)) break; for(i=1;i<=n;i++) scanf("%lld",&a[i]); for(i=1;i<=m;i++) scanf("%lld",&b[i]); p=b[1]; for(i=2;i<=m;i++) { k=gcd(p,b[i]); if (p/k>BG/b[i]) p=h+1; else p=p/k*b[i]; } ans=0; calc(1,0,1); printf("%lld/n",ans); } }