Codeforces Round #179 (Div. 2) B. Yaroslav and Two Strings 【组合数学】
2013-04-12 02:56
531 查看
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]。
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; }
相关文章推荐
- Codeforces Round #179 (Div. 2) B. Yaroslav and Two Strings(容斥原理)
- Codeforces Round #179 (Div. 2) B. Yaroslav and Two Strings
- Codeforces Round #179 (Div. 2) B.Yaroslav and Two Strings(容斥原理)
- Codeforces Round #179 (Div. 2) B. Yaroslav and Two Strings (容斥原理)
- Codeforces Round #179 (Div. 2) B (codeforces 296b) Yaroslav and Two Strings
- codeforces round 404 div2 D Anton and School - 2 组合数学
- B. Yaroslav and Two Strings
- Codeforces Round #177 (Div. 2)---D. Polo the Penguin and Houses (组合数学+暴力)
- Codeforces Round #447 (Div. 2) B. Ralph And His Magic Field(数论,组合数学)
- Codeforces Round #419 (Div. 2) (Codeforces 815B) D. Karen and Test 组合数学
- CodeForces Round #179 (296B) - Yaroslav and Two Strings
- CodeForces 296 B.Yaroslav and Two Strings(dp)
- 4000 codeforces 296B - Yaroslav and Two Strings (DP+容斥)
- Codeforces Round #167 (Div. 2) D. Dima and Two Sequences 排列组合
- Codeforces Round #404(Div. 2)D. Anton and School - 2【组合数学+思维】好题!好题!
- Codeforces Round #272 (Div. 2) B. Dreamoon and WiFi dp,组合数学
- CodeForces 272 D.Dima and Two Sequences(组合数学)
- Codeforces Round #404 (Div. 2) D. Anton and School - 2(组合数学)
- Codeforces Round #177 (Div. 1) B. Polo the Penguin and Houses【组合数学】
- 情况动态规划CodeForces Round #179 (296B) - Yaroslav and Two Strings