UVA 10458 - Cricket Ranking(容斥原理)

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;}