Segment---hdu5666 (two long long to find remainder of long long)

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5666

Test instructions: There is a line x+y=n, to find the line and the axis of the triangle enclosed by how many integer points, not on the axis and line, so that the result of the P-remainder

The obvious answer is (1+2+3+...+a-2)%P = ((A-1) (a-2)/2)%P;

Because the range of a and p are relatively large, the direct multiplication will explode ll's, so you can use the idea of a fast power matrix;

X*y = 2*X*Y/2; Y is an even number;

X*y = x + 2*x*y/2; Y is odd;

So there's a bit of code:

#include <iostream> #include <vector> #include <stdio.h> #include <string.h> #include < stdlib.h> #include <algorithm> #include <math.h>using namespace std; #define N 1050#define PI 4*atan (1) # Define met (A, b) memset (A, B, sizeof (a)) typedef long Long LL; ll Mul (ll X, ll y, ll Mod) {    if (y = = 0) return 0;    LL ans = 2 * Mul (x, Y/2, Mod)%mod;    if (y%2)        ans = (ans+x)%mod;    return ans;} int main () {    int T;    scanf ("%d", &t);    while (t--)    {        LL ans, p, q, N, mod;        scanf ("%i64d%i64d", &n, &mod);        p = n-1; Q = n-2;        if (p&1)            ans = Mul (p, Q/2, mod);        else            ans = Mul (P/2, q, mod);        printf ("%i64d\n", ans);    }    return 0;}

　　

Segment---hdu5666 (two long long to find remainder of long long)