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UVA 10458 - Cricket Ranking
题目链接
题意：给定k个区间，要求用这些数字范围去组合成n，问有几种组合方式
思路：容斥原理，容斥是这样做：已知n个组成s，不限值个数的话，用隔板法求出情况为C(s + n - 1, n - 1)，但是这部分包含了超过了，那么就利用二进制枚举出哪些是超过的，实现把s减去f(i) + 1这样就保证这个位置是超过的，减去这部分后，有多减的在加回来，这就满足了容斥原理的公式，个数为奇数的时候减去，偶数的时候加回
代码：
#include <cstdio>
#include <cstring>
#include <iostream>using namespace std;
typedef long long ll;
const int MAXN = 1005;struct bign {int len;ll num[MAXN];bign () {len = 0;memset(num, 0, sizeof(num));}bign (ll number) {*this = number;}bign (const char* number) {*this = number;}void DelZero ();void Put ();void operator = (ll number);void operator = (char* number);bool operator < (const bign& b) const;bool operator > (const bign& b) const { return b < *this; }bool operator <= (const bign& b) const { return !(b < *this); }bool operator >= (const bign& b) const { return !(*this < b); }bool operator != (const bign& b) const { return b < *this || *this < b;}bool operator == (const bign& b) const { return !(b != *this); }void operator ++ ();void operator -- ();bign operator + (const int& b);bign operator + (const bign& b);bign operator - (const int& b);bign operator - (const bign& b);bign operator * (const ll& b);bign operator * (const bign& b);bign operator / (const ll& b);//bign operator / (const bign& b);int operator % (const int& b);
};/*Code*/int k;
long long n, f[10];int bitcount(int x) {return x == 0 ? 0 : bitcount(x>>1) + (x&1);
}bign C(long long n, long long m) {bign ans = 1;for (long long i = 0; i < m; i++)ans = ans * (n - i) / (i + 1);return ans;
}int main() {while (~scanf("%d%lld", &k, &n)) {long long l, r;for (int i = 0; i < k; i++) {scanf("%lld%lld", &l, &r);f[i] = r - l;n -= l;}bign ans1 = 0LL, ans2 = 0LL;for (int i = 0; i < (1<<k); i++) {long long s = n;for (int j = 0; j < k; j++) {if (i&(1<<j)) {s -= f[j] + 1;if (s < 0) break;}}if (s < 0) continue;if (bitcount(i)&1) ans2 = ans2 + C(s + k - 1, k - 1);else ans1 = ans1 + C(s + k - 1, k - 1);}(ans1 - ans2).Put();printf("\n");}return 0;
}/*********************************************/void bign::DelZero () {while (len && num[len-1] == 0)len--;if (len == 0) {num[len++] = 0;}
}void bign::Put () {for (int i = len-1; i >= 0; i--) printf("%lld", num[i]);
}void bign::operator = (char* number) {len = strlen (number);for (int i = 0; i < len; i++)num[i] = number[len-i-1] - '0';DelZero ();
}void bign::operator = (ll number) {len = 0;while (number) {num[len++] = number%10;number /= 10;}DelZero ();
}bool bign::operator < (const bign& b) const {if (len != b.len)return len < b.len;for (int i = len-1; i >= 0; i--)if (num[i] != b.num[i])return num[i] < b.num[i];return false;
}void bign::operator ++ () {int s = 1;for (int i = 0; i < len; i++) {s = s + num[i];num[i] = s % 10;s /= 10;if (!s) break;}while (s) {num[len++] = s%10;s /= 10;}
}void bign::operator -- () {if (num[0] == 0 && len == 1) return;int s = -1;for (int i = 0; i < len; i++) {s = s + num[i];num[i] = (s + 10) % 10;if (s >= 0) break;}DelZero ();
}bign bign::operator + (const int& b) {bign a = b;return *this + a;
}bign bign::operator + (const bign& b) {int bignSum = 0;bign ans;for (int i = 0; i < len || i < b.len; i++) {if (i < len) bignSum += num[i];if (i < b.len) bignSum += b.num[i];ans.num[ans.len++] = bignSum % 10;bignSum /= 10;}while (bignSum) {ans.num[ans.len++] = bignSum % 10;bignSum /= 10;}return ans;
}bign bign::operator - (const int& b) {bign a = b;return *this - a;
}bign bign::operator - (const bign& b) {ll bignSub = 0;bign ans;for (int i = 0; i < len || i < b.len; i++) {bignSub += num[i];if (i < b.len)bignSub -= b.num[i];ans.num[ans.len++] = (bignSub + 10) % 10;if (bignSub < 0) bignSub = -1;else bignSub = 0;}ans.DelZero();return ans;
}bign bign::operator * (const ll& b) {ll bignSum = 0;bign ans;ans.len = len;for (int i = 0; i < len; i++) {bignSum += num[i] * b;ans.num[i] = bignSum % 10;bignSum /= 10;}while (bignSum) {ans.num[ans.len++] = bignSum % 10;bignSum /= 10;}return ans;
}bign bign::operator * (const bign& b) {bign ans;ans.len = 0; for (int i = 0; i < len; i++){ int bignSum = 0; for (int j = 0; j < b.len; j++){ bignSum += num[i] * b.num[j] + ans.num[i+j]; ans.num[i+j] = bignSum % 10; bignSum /= 10;} ans.len = i + b.len; while (bignSum){ ans.num[ans.len++] = bignSum % 10; bignSum /= 10;} } return ans;
}bign bign::operator / (const ll& b) {bign ans;ll s = 0;for (int i = len-1; i >= 0; i--) {s = s * 10 + num[i];ans.num[i] = s/b;s %= b;}ans.len = len;ans.DelZero();return ans;
}int bign::operator % (const int& b) {bign ans;int s = 0;for (int i = len-1; i >= 0; i--) {s = s * 10 + num[i];ans.num[i] = s/b;s %= b;}return s;
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