cf1056B. Divide Candies(数论 剩余系)

题意

题目链接

求满足\(i^2 + j^2 \% m = 0\)的数对\((i, j)\)的个数，\(1 \leqslant i, j \leqslant 10^9, m \leqslant 1000\)

sol

发这篇博客的目的就是为了证明一下我到底有多菜。

mdzz小学组水题我想了40min都没想出来。这要是出在noip 2019的话估计我又要做不出day1t1了。。

\(i^2 + j^2 \% m = i \% m * i \% m + j \% m * j \% m\)

枚举剩余系即可

此题完结

/*
*/
#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long 
#define ll long long 
#define rg register 
#define pt(x) printf("%d ", x);
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 1e6 + 10, inf = 1e9 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m, vis[3001][3001];
int get(int l, int r) {
    return (r - l) / m + (l > 0);
}
main() {
    n = read(); m = read();
    int ans = 0;
    for(int i = 1; i <= m; i++) {
        for(int j = 1; j <= m; j++) {
            if((i * i + j * j) % m == 0) {
                vis[i % m][j % m] = 1;
            }
        }
    }
    for(int i = 0; i <= min(n, m); i++)
        for(int j = 0; j <= min(n, m); j++)
            if(vis[i][j] && ((i * i + j * j) % m == 0))
                (ans += get(i, n) * get(j, n)); 
    cout << ans;
    return 0;   
}
/*

*/