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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;
}

UVA 10458 - Cricket Ranking(容斥原理),布布扣,bubuko.com

UVA 10458 - Cricket Ranking(容斥原理)

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原文地址：http://blog.csdn.net/accelerator_/article/details/38182349