uva 10458 - Cricket Ranking(容斥+高精度)

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题目连接：uva 10458 - Cricket Ranking

题目大意：给定k和n，表示有k个比赛，总共要的n分，每个比赛可以得l~r的分数，问说可以有多少种得分方式。

解题思路：容斥，可以参考Codeforces 451E.

#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*/
const int maxn = 10;
ll k, n, s, l[maxn], r[maxn];

void init () {
    s = 0;
    for (int i = 0; i < k; i++) {
        scanf("%lld%lld", &l[i], &r[i]);
        r[i] -= l[i];
        n -= l[i];
        s += r[i];
    }
}

bign C (ll u, ll v) {

    v = min(v, u-v);

    bign ans = 1;
    for (ll i = 0; i < v; i++)
        ans = ans * (u-i) / (i+1);
    return ans;
}

void solve () {
    if (n < 0 || n > s) {
        printf("0\n");
        return ;
    }

    bign add = 0LL, del = 0LL;

    for (int i = 0; i < (1<<k); i++) {
        ll sum = 0, cnt = 0;
        for (int j = 0; j < k; j++) {
            if (i&(1<<j)) {
                cnt++;
                sum += (r[j] + 1);
            }
        }

        if (n < sum)
            continue;

        if (cnt&1)
            del = del + C(n-sum+k-1, k-1);
        else
            add = add + C(n-sum+k-1, k-1);
    }

    if (add < del) {
        printf("0\n");
        printf("\n");
    }

    /*
    add.Put();
    printf("\n");
    del.Put();
    printf("\n");
    */
    bign ans = add - del;
    bign rec = ans + del;

    while (add != rec);

    ans.Put();
    printf("\n");
}

int main () {
    while (scanf("%lld%lld", &k, &n) == 2) {
        init ();
        solve();
    }
    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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uva 10458 - Cricket Ranking(容斥+高精度)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/38173167