Codeforces Round #179 (Div. 2) B. Yaroslav and Two Strings 【组合数学】

http://codeforces.com/contest/296/problem/B

对于两个字符串ch1和ch2,开四个数组a[i],b[i],c[i],d[i]分别表示 所有的情况数 、ch1[i]<=ch2[i]的情况数、ch1[i]>=ch2[i]的情况数、ch1[i]==ch2[i]的情况数,那么根据容斥原理,有ans = ∏a[i] - ∏b[i] - ∏c[i] + ∏d[i]。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <queue>
using namespace std;
template <class T> void checkmin(T &t,T x) {if(x < t) t = x;}
template <class T> void checkmax(T &t,T x) {if(x > t) t = x;}
template <class T> void _checkmin(T &t,T x) {if(t==-1) t = x; if(x < t) t = x;}
template <class T> void _checkmax(T &t,T x) {if(t==-1) t = x; if(x > t) t = x;}
typedef pair <int,int> PII;
typedef pair <double,double> PDD;
typedef long long ll;
#define foreach(it,v) for(__typeof((v).begin()) it = (v).begin(); it != (v).end ; it ++)
#define MOD 1000000007
int n ;
ll a[101000] , b[101000] , c[101000] , d[101000];
char ch1[101000] , ch2[101000];

void gcd(ll a , ll b , ll &d , ll &x , ll &y) {
    if(!b) {d = a; x = 1; y = 0;}
    else { gcd(b , a%b,d,y , x); y -= x * (a/b); }
}
ll inv(ll a , ll n) {
    ll d , x , y;
    gcd(a , n , d,  x , y);
    return d == 1 ? (x+n)%n : -1;
}

void debug() {
    for(int i=0;i<n;i++) cout << a[i] << " ";
    cout << endl;
    for(int i=0;i<n;i++) cout << b[i] << " ";
    cout << endl;
    for(int i=0;i<n;i++) cout << c[i] << " ";
    cout << endl;
    for(int i=0;i<n;i++) cout << d[i] << " ";
    cout << endl;
}

int main() {
    cin >> n;
    scanf("%s%s",ch1,ch2);
    for(int i=0;i<n;i++) {
        if(ch1[i] == '?' && ch2[i] == '?') { a[i] = b[i] = 55;c[i] = 10; d[i] = 100; }
        else if(ch1[i] == '?') { a[i] = ch2[i] - '0'+1; b[i] = 11-a[i]; c[i] = 1; d[i] = 10; }
        else if(ch2[i] == '?') { b[i] = ch1[i] - '0'+1; a[i] = 11-b[i]; c[i] = 1; d[i] = 10; }
        else {
            if(ch1[i] <= ch2[i]) a[i] = 1;
            if(ch1[i] >= ch2[i]) b[i] = 1;
            if(ch1[i] == ch2[i]) c[i] = 1;
            d[i] = 1;
        }
    }
    ll a1 = 1 , a2 = 1 , a3 = 1 , a4 = 1;
        for(int i=0;i<n;i++) {
            a1 *= a[i]; a1 %= MOD;
            a2 *= b[i]; a2 %= MOD;
            a3 *= c[i]; a3 %= MOD;
            a4 *= d[i]; a4 %= MOD;
        }
        ll ans = (a4-a1-a2+a3) % MOD;
        if(ans < 0) ans += MOD;
        cout << ans << endl;
        //debug();
    return 0;
}

 

posted @ 2013-04-12 02:56  aiiYuu  阅读(...)  评论(...编辑  收藏