标签:
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4857 Accepted Submission(s):
1611
1 #define M 15 2 #include<iostream> 3 using namespace std; 4 #include<cstdio> 5 #include<cstring> 6 #define ll long long 7 void exgcd(ll a,ll b,ll &x,ll &y,ll &gcd) 8 { 9 if(b==0) 10 { 11 x=1;y=0; 12 gcd=a; 13 return; 14 } 15 exgcd(b,a%b,x,y,gcd); 16 ll t=x; 17 x=y; 18 y=t-(a/b)*y; 19 } 20 int main() 21 { 22 int test; 23 scanf("%d",&test); 24 while(test--) 25 { 26 int n,m; 27 scanf("%d%d",&n,&m); 28 ll a[M],b[M]; 29 ll gcd,a1,a2,b1,b2,x,y; 30 for(int i=0;i<m;++i) 31 scanf("%I64d",&a[i]);/*我犯的一个不明显的小错误,a,b数组未赋初值,而且scanf(“%d)读入,结果,a,b的数组中是非常奇怪的数,所以以后定义数组一定要初始化*/ 32 for(int i=0;i<m;++i) 33 scanf("%I64d",&b[i]); 34 a1=a[0];b1=b[0]; 35 bool flag=false; 36 for(int i=1;i<m;++i) 37 { 38 a2=a[i];b2=b[i]; 39 exgcd(a1,a2,x,y,gcd); 40 if((b2-b1)%gcd) 41 { 42 flag=true; 43 break; 44 } 45 ll t=a2/gcd; 46 x=(x*(b2-b1))/gcd; 47 x=(x%t+t)%t; 48 b1=a1*x+b1; 49 a1=(a1*a2)/gcd; 50 b1=(b1%a1+a1)%a1; 51 } 52 if(flag||n<b1) 53 { 54 printf("0\n"); 55 } 56 else cout<<((n-b1)/a1+1-(b1==0?1:0))<<endl; 57 } 58 return 0; 59 }
标签:
原文地址:http://www.cnblogs.com/c1299401227/p/5514084.html