ZOJ4036 Lucky 7

Posted 2021-12-18 misuchii

给出n个数和一个数b 问这n个数中是否存在一个数加上b之后能被7整除 如果存在的话 即是幸运的

由(a + b) % c = a % c + b % c可知 先用b对7求余 再用n个数分别对7求余 如果这n个数%7的值加上b等于7或者b和这n个数%7的值都为0 就满足题意了

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 typedef unsigned long long ull;
 5 
 6 namespace io 
 7     const int SIZE = 1e7 + 10;
 8     char inbuff[SIZE];
 9     char *l, *r;
10     inline void init() 
11         l = inbuff;
12         r = inbuff + fread(inbuff, 1, SIZE, stdin);
13     
14     inline char gc() 
15         if (l == r) init();
16         return (l != r) ? *(l++) : EOF;
17     
18     void read(int &x) 
19         x = 0; char ch = gc();
20         while(!isdigit(ch)) ch = gc();
21         while(isdigit(ch)) x = x * 10 + ch - ‘0‘, ch = gc();
22     
23  using io::read;
24 
25 int main()
26     int t;
27     scanf("%d", &t);
28     int n, b;
29     int x;
30     bool flag;
31     while (t--)
32         scanf("%d%d", &n, &b);
33         b = b % 7;
34         flag = false;
35         for (int i = 0; i < n; i++)
36             cin>>x;
37             x = x % 7;
38             if (x + b == 7 || (!x && !b)) flag = true;
39         
40         if (flag) cout<<"Yes"<<endl;
41         else cout<<"No"<<endl;
42     
43     return 0;
44