HDU-1573 X问题(中国剩余定理)

Posted 2022-09-16 MichaelZona

X问题

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5521    Accepted Submission(s): 1875

Problem Description

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)。

 

 

Input

输入数据的第一行为一个正整数T，表示有T组测试数据。每组测试数据的第一行为两个正整数N，M (0 < N <= 1000,000,000 , 0 < M <= 10)，表示X小于等于N，数组a和b中各有M个元素。接下来两行，每行各有M个正整数，分别为a和b中的元素。

 

 

Output

对应每一组输入，在独立一行中输出一个正整数，表示满足条件的X的个数。

 

 

Sample Input

3 10 3 1 2 3 0 1 2 100 7 3 4 5 6 7 8 9 1 2 3 4 5 6 7 10000 10 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9

 

 

Sample Output

1 0 3

 

 

Author

lwg

 

 

Source

​​HDU 2007-1 Programming Contest ​​

 

 

Recommend

linle

 

 

此题乃中国剩余定理 解模线性方程组的模板题。

 

1 #include "bits/stdc++.h"
 2 using namespace std;
 3 typedef long long LL;
 4 const int MAX=15;
 5 LL cas;
 6 LL n,m;
 7 LL aa[MAX],r[MAX],lcm;
 8 LL gcd(LL a,LL b)
 9     if (b==0)return a;
10     return gcd(b,a%b);
11 
12 LL exgcd(LL a,LL b,LL &x,LL &y)
13     if (b==0)x=1,y=0;return a;
14     LL d=exgcd(b,a%b,x,y),t=x;x=y,y=t-(a/b)*y;
15     return d;
16 
17 LL modeqset()
18     LL i,j;
19     LL a,b,d,c,x,y,t;
20     for (i=2;i<=n;i++)
21         a=aa[i-1],b=aa[i];
22         c=r[i]-r[i-1];
23         d=exgcd(a,b,x,y);
24         if (c%d) return -1;
25         t=aa[i]/d;
26         x=(x*(c/d)%t+t)%t;
27         r[i]=r[i-1]+aa[i-1]*x;
28         aa[i]=aa[i-1]/d*aa[i];
29     
30     return r[n];
31 
32 int main()
33     freopen ("x.in","r",stdin);
34     freopen ("x.out","w",stdout);
35     LL i,j;
36     scanf("%lld",&cas);
37     while (cas--)
38         scanf("%lld%lld",&m,&n);
39         lcm=1;
40         for (i=1;i<=n;i++)
41             scanf("%lld",aa+i);
42             lcm=lcm/gcd(lcm,aa[i])*aa[i];
43         
44         for (i=1;i<=n;i++)
45          scanf("%lld",r+i);
46         LL ans=modeqset();
47         if (ans==-1 || ans>m)
48             puts("0");
49             continue;
50         
51         if (ans==0)
52          printf("%lld\\n",(m-ans)/lcm);
53         else
54          printf("%lld\\n",(m-ans)/lcm+1);
55     
56     return 0;
57