HDU 1573 X问题 中国剩余定理
链接:http://acm.hdu.edu.cn/showproblem.php?
题意:求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod a[i] = b[i], … (0 < a[i] <= 10)。
思路:中国剩余定理的模板题(全部除数相互不互质版),假设找不到这种数或者最小的X大于N。输出零。
资料:http://yzmduncan.iteye.com/blog/1323599/
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <map> #include <cstdlib> #include <queue> #include <stack> #include <vector> #include <ctype.h> #include <algorithm> #include <string> #include <set> #define PI acos(-1.0) #define maxn 10005 #define INF 0x7fffffff #define eps 1e-8 typedef long long LL; typedef unsigned long long ULL; using namespace std; LL MOD; LL extend_gcd(LL a, LL b, LL &x, LL &y) { if(b==0) { x=1; y=0; return a; } LL r=extend_gcd(b,a%b,x,y); LL t=x; x=y; y=t-a/b*y; return r; } LL inv(LL a,LL m) { LL d,x,y; d=extend_gcd(a,m,x,y); if (d==1) { x=(x%m+m)%m; return x; } else return -1; } LL gcd(LL a,LL b) { return b==0?a:gcd(b,a%b); } bool merge(LL a1,LL m1,LL a2,LL m2,LL &a3,LL &m3) { LL d=gcd(m1,m2); LL c=a2-a1; if(c%d) return false; c=(c%m2+m2)%m2; c/=d; m1/=d; m2/=d; c*=inv(m1,m2); c%=m2; c*=m1*d; c+=a1; m3=m1*m2*d; a3=(c%m3+m3)%m3; return true; } LL CRT_next(LL a[],LL m[],int n) { LL a1=a[0],m1=m[0],a2,m2; for(int i=1;i<n;i++) { LL aa,mm; a2=a[i],m2=m[i]; if(!merge(a1,m1,a2,m2,aa,mm)) return -1; a1=aa; m1=mm; } MOD=m1; LL aa=(a1%m1+m1)%m1; if(aa==0) aa+=m1; return aa; } int main() { int T; LL a[55],b[55]; scanf("%d",&T); for(int ii=1; ii<=T; ii++) { int tot; LL t1; scanf("%I64d%d",&t1,&tot); for(int i=0; i<tot; i++) scanf("%I64d",&a[i]); for(int i=0; i<tot; i++) scanf("%I64d",&b[i]); if(tot==1) { if(b[0]==0) b[0]+=a[0]; if(t1<b[0]) printf("0\n"); else printf("%I64d\n",(t1-b[0])/a[0]+1); } else { LL ans=CRT_next(b,a,tot); if(ans==-1) printf("0\n"); else if(ans>t1) printf("0\n"); else printf("%I64d\n",(t1-ans)/MOD+1); } } return 0; }