Ultraviolet A 10542,

Ultraviolet A 10542,

Link: Ultraviolet A 10542-Hyper-drive

Given a line segment in an n-dimensional space, the line segment goes through several grids.

Solution: For a line segment, you can move one point to the origin point and change it
(0, 0) to (a, B) This line segment, take two-dimensional as an example, each time will move from a grid to another grid, can be x + 1 coordinates, can be y + 1, so the total should be a + b-1, deduct x + 1, y + 1 case gcd (a, B)-1 (origin ). The calculation result of ing to n-dimensional data is calculated using the exclusion principle.

/*********************** * (0, 0, 0, ...) -> (a, b, c, ...) * ans = a + b + c +.. - gcd(a,b) - gcd(a,c) - .. + gcd(a, b, c) ...***********************/#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 15;int N, d[maxn], g[2][maxn];inline int gcd (int a, int b) {    return b == 0 ? a : gcd (b, a%b);}void init () {    scanf("%d", &N);    for (int i = 0; i < 2; i++)        for (int j = 0; j < N; j++)            scanf("%d", &g[i][j]);    for (int i = 0; i < N; i++)        d[i] = abs(g[0][i] - g[1][i]);}ll solve () {    ll ans = 0;    for (int i = 1; i < (1<<N); i++) {        int sign = -1, tmp = -1;        for (int j = 0; j < N; j++) {            if (i&(1<<j)) {                tmp = (tmp == -1 ? d[j] : gcd(tmp, d[j]));                sign *= -1;            }        }        ans += tmp * sign;    }    return ans;}int main () {    int cas;    scanf("%d", &cas);    for (int i = 1; i <= cas; i++) {        init();        printf("Case %d: %lld\n", i, solve());    }    return 0;}