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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;
}
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