UVA 10542 - Hyper-drive（容斥原理）

UVA 10542 - Hyper-drive

题目链接

题意：给定一些个d维的方块，给定两点，求穿过多少方块

思路：容斥原理，每次选出一些维度，如果gcd(a, b)，就会穿过多少点，对应的就减少穿过多少方块，所以最后得到式子d1 + d2 + .. dn - gcd(d1, d2)..+gcd(d1, d2, d3)...

代码：

#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;}