NOIp 2014解方程

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

#include<iostream>
#include<cstdio>

using namespace std;

typedef long long ll;

const int mod = 1e9+7;

ll read(){
    ll ans = 0;char ch = getchar(),last = ch;
    while(ch < '0'||ch > '9')last = ch,ch = getchar();
    while('0' <= ch&&ch <= '9')ans = (ans*10+ch-'0')%mod,ch = getchar();
    if(last == '-')return -ans;return ans;
}

int a[110],sta[110];
int n,m,top = 0;

bool check(int x){
    ll tot = 0;
    for(int i = n;i >= 0;i--)tot = (tot*x + a[i])%mod;
    return !tot;
}

int main(){
    n = read(),m = read();
    for(int i = 0;i <= n;i++)a[i] = read();
    for(int i = 1;i <= m;i++)if(check(i))sta[++top] = i;
    printf("%d
",top);
    for(int i = 1;i <= top;i++)printf("%d
",sta[i]);    
return 0;
}