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
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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 }