Sha 10542-hyper-drive (Anti-DDoS principle)

Ultraviolet A 10542-hyper-drive

Question Link

Question: given some D-dimensional squares, given two points, how many squares have to pass through

Idea: refresh principle: select some dimensions each time. If gcd (a, B) goes through the number of points, the corresponding number of blocks is reduced, so the formula D1 + D2 + is obtained .. DN-gcd (D1, D2 ).. + gcd (D1, D2, D3 )...

Code:

#include <cstdio>#include <cstring>typedef long long ll;int t, d, a[15];int bitcount(int x) {int ans = 0;while (x) {ans += (x&1);x >>= 1;}return ans;}int gcd(int a, int b) {while (b) {int tmp = a % b;a = b;b = tmp;}return a;}int main() {int cas = 0;scanf("%d", &t);while (t--) {scanf("%d", &d);for (int i = 0; i < d; i++)scanf("%d", &a[i]);int tmp;for (int i = 0; i < d; i++) {scanf("%d", &tmp);a[i] -= tmp;if (a[i] < 0) a[i] = -a[i];}ll ans = 0;for (int i = 0; i < (1<<d); i++) {int cnt = bitcount(i);int sum = 0;for (int j = 0; j < d; j++) {if (i&(1<<j))sum = gcd(sum, a[j]);}if (cnt&1) ans += sum;else ans -= sum;}printf("Case %d: %lld\n", ++cas, ans);}return 0;}

Sha 10542-hyper-drive (Anti-DDoS principle)