乍看一眼以为是高精，随便写了个usigned long long交上去拿了50分，翻了翻题解才知道用了我没用过的算法——秦九韶算法，算是道模板了吧，不过这道题中的对10^10000的数的处理值得记住，并不是告诉了模一个质数才是模质数的题，凡是位数特别多且不要求求具体答案的大概都可能用到模质数吧
代码
1 #include<iostream> 2 #include<cstdio> 3 4 using namespace std; 5 6 typedef long long ll; 7 8 const int mod = 1e9+7; 9 10 ll read(){ 11 ll ans = 0;char ch = getchar(),last = ch; 12 while(ch < ‘0‘||ch > ‘9‘)last = ch,ch = getchar(); 13 while(‘0‘ <= ch&&ch <= ‘9‘)ans = (ans*10+ch-‘0‘)%mod,ch = getchar(); 14 if(last == ‘-‘)return -ans;return ans; 15 } 16 17 int a[110],sta[110]; 18 int n,m,top = 0; 19 20 bool check(int x){ 21 ll tot = 0; 22 for(int i = n;i >= 0;i--)tot = (tot*x + a[i])%mod; 23 return !tot; 24 } 25 26 int main(){ 27 n = read(),m = read(); 28 for(int i = 0;i <= n;i++)a[i] = read(); 29 for(int i = 1;i <= m;i++)if(check(i))sta[++top] = i; 30 printf("%d\n",top); 31 for(int i = 1;i <= top;i++)printf("%d\n",sta[i]); 32 return 0; 33 }
原文：https://www.cnblogs.com/Wangsheng5/p/11530956.html