题意：

给你一个数列：

f(0) = 0

f(n) = g(n,f(n-1))

g(x,y) = ((y-1)*x^5+x^3-xy+3x+7y)%9973

让你求f(n)  n <= 1e8

思路：

令m = 9973

容易观察g(x,y) = g(x%m,y)

f(x+m) = g( (x+m) %m , f(x+m-1))........

可以得到 f(x+m) = (A*f(x)+B)%m

f(x+2m) = (A*f(x+m)+B)%m

,.....

令x+km = n

先求出f(x) 在求出A,B然后算出f(n)

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int m = 9973;
int A,B;
int n;
int x,cnt;
int getX5(int x){
    int ret = 1;
    for(int i = 1; i <= 5; i++) {
        ret = (x*ret)%m;
    }
    return ret;
}
int getX3(int x) {
    int ret = 1;
    for(int i = 1; i <= 3; i++) {
        ret = (ret*x)%m;
    }
    return ret;
}
int getPos(int x) {
    return (x%m+m)%m;
}
int func(int n) {
    if(n==0) return 0;
    return getPos((getPos(getX5(n)-n+7)*func(n-1))%m+getPos((-getX5(n)+getX3(n)+3*n)));

}
void solve() {
    A = 1;
    B = 0;
    int ret = func(x);
    for(int i = x+m; i >= x+1; i--) {
        int k = i%m;
        B = (B+A*getPos((-getX5(k)+getX3(k)+3*k)))%m;
        A = (A*getPos(getX5(k)-k+7))%m;
    }
    for(int i = 1; i <= cnt; i++) {
        ret = (A*ret+B)%m;
    }
    printf("%d\n",ret);


}
int main() {
    while(~scanf("%d",&n)) {
        x = n%m;
        cnt = n/m;
        solve();

    }
    return 0;
}

Ural 1309 Dispute (递推)