【AtCoder】ExaWizards 2019

ExaWizards 2019

C - Snuke the Wizard

发现符文的相对位置不变，直接二分某个位置是否到达最左或最右来计算

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 +c - '0';
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}

int N,Q;
char s[MAXN];
pii op[MAXN];
int getplace(int p) {
    for(int i = 1 ; i <= Q ; ++i) {
        if(s[p] == 'A' + op[i].fi) {
            p += op[i].se;
        }
    }
    return p;
}
int findL() {
    int L = 0,R = N;
    while(L < R) {
        int mid = (L + R + 1) >> 1;
        if(getplace(mid) == 0) L = mid;
        else R = mid - 1;
    }
    return L;
}
int findR() {
    int L = 1,R = N + 1;
    while(L < R) {
        int mid = (L + R) >> 1;
        if(getplace(mid) == N + 1) R = mid;
        else L = mid + 1;
    }
    return L;
}
void Solve() {
    read(N);read(Q);
    scanf("%s",s + 1);
    char t[5],d[5];
    for(int i = 1 ; i <= Q ; ++i) {
        scanf("%s%s",t + 1,d + 1);
        int a,b;
        if(d[1] == 'L') b = -1;
        else b = 1;
        a = t[1] - 'A';
        op[i] = mp(a,b);
    }
    int L = findL(),R = findR();
    out(R - L - 1);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

D - Modulo Operations

如果按照我们对值产生贡献的序列一定是一个递减的序列

我们只要按照值从大到小排序后记录\(dp[i][j]\)为到第i个数取模的数是j的概率是多少

转移的时候如果对当前值进行取模，则乘上一个\(\frac{1}{N - i + 1}\)

否则取乘一个\(\frac{N- i}{N - i + 1}\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 +c - '0';
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N,X,a[205],fac[205],inv[205];
int dp[205][100005];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
void Solve() {
    read(N);read(X);
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    sort(a + 1,a + N + 1,[](int a,int b){return a > b;});
    fac[0] = 1;
    for(int i = 1 ; i <= N ; ++i) fac[i] = mul(fac[i - 1],i);
    inv[1] = 1;
    for(int i = 2 ; i <= N ; ++i) inv[i] = mul(inv[MOD % i],MOD - MOD / i);
    dp[0][X] = 1;
    for(int i = 1 ; i <= N ; ++i) {
        for(int j = 0 ; j <= X ; ++j) {
            if(j < a[i]) update(dp[i][j],dp[i - 1][j]);
            else {
                update(dp[i][j % a[i]],mul(dp[i - 1][j],inv[N - i + 1]));
                update(dp[i][j],mul(dp[i - 1][j],mul(N - i,inv[N - i + 1])));
            }
        }
    }
    int ans = 0;
    for(int j = 0 ; j <= X ; ++j) {
        update(ans,mul(dp[N][j],j));
    }
    ans = mul(ans,fac[N]);
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

E - Black or White

我们把点画在坐标轴上,每个i相当于画一条斜线，那么如果这个斜线上的点都有让B减少的一步，那么答案就是\(\frac{1}{2}\)

否则处理出剩了多少个白点，和剩了多少个黑点的情况，每条斜线最多涉及两个这样的点，把他们的贡献加上或减掉即可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 +c - '0';
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int B,W;
int fac[MAXN],invfac[MAXN],pw[MAXN];
int r[MAXN],c[MAXN];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int C(int n,int m) {
    if(n < m) return 0;
    return mul(fac[n],mul(invfac[m],invfac[n - m]));
}

int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
        if(c & 1) res = mul(res,t);
        t = mul(t,t);
        c >>= 1;
    }
    return res;
}
void Solve() {
    read(B);read(W);
    fac[0] = 1;
    for(int i = 1 ; i <= B + W ; ++i) fac[i] = mul(fac[i - 1],i);
    invfac[B + W] = fpow(fac[B + W],MOD - 2);
    for(int i = B + W - 1 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
    pw[0] = 1;pw[1] = invfac[2];
    for(int i = 2 ; i <= B + W ; ++i) pw[i] = mul(pw[i - 1],pw[1]);
    r[0] = pw[W];
    for(int i = 1; i <= B ; ++i) {
        r[i] = inc(r[i - 1],mul(pw[1],mul(pw[i + W - 1],C(i + W - 1,i))));
    }
    c[0] = pw[B];
    for(int i = 1 ; i <= W ; ++i) {
        c[i] = inc(c[i - 1],mul(pw[1],mul(pw[B - 1 + i],C(B - 1 + i,i))));
    }
    for(int i = 0 ; i < B + W ; ++i) {
        int res = (MOD + 1) / 2;
        if(i >= B) {
            res = inc(res,MOD - mul((MOD + 1) / 2,c[i - B]));
        }
        if(i >= W) {
            res = inc(res,mul((MOD + 1) / 2,r[i - W]));
        }
        out(res);enter;
    }

}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}