XJOJ 提高组2

T1 Strings

也不是特别难的一道题，主要还是考查自己的细心，求一下乘法逆元再加上一些简单的组合就搞定了。

/*    ID:Agreement*///invincible#include <bits/stdc++.h>#define rep( i , l , r ) for( int i = (l) ; i <= (r) ; ++i )#define per( i , r , l ) for( int i = (r) ; i >= (l) ; --i )#define erep( i , u ) for( int i = head[(u)] ; ~i ; i = e[i].nxt )using namespace std;typedef long long ll;const int maxn = 1e5 + 5 , MOD = 1e9 + 7;int N , K;char s[maxn] , t[maxn];ll fac[maxn] , inv[maxn] , MOD25[maxn];ll _pw( ll x , ll y ){    ll t = 1ll;    for( ; y ; y >>= 1 ){        if( y & 1 ) t = t * x % MOD;        x = x * x % MOD;    }    return t;}ll prework(){    fac[0] = 1;    rep( i , 1 , N ) fac[i] = fac[i - 1] * (ll)i % MOD;    inv[N] = _pw( fac[N] , MOD - 2 );    per( i , N - 1 , 0 ) inv[i] = inv[i + 1] * (ll)(i + 1ll) % MOD;    MOD25[0] = 1;    rep( i , 1 , N ) MOD25[i] = MOD25[i - 1] * 25 % MOD;}ll C( int n , int r ){    return fac[n] * inv[n - r] % MOD * inv[r] % MOD;}ll solve( char now[] , char sta[] ){    ll res = 1 ; int diff = K , t = 0;    rep( i , 1 , N  ){        for( char c = 'a' ; c <= (now[i - 1] - 1) ; ++c ){            if( sta[i - 1] != c ) t = diff - 1;            else t = diff;            res = ( res + C( N - i , t ) * MOD25[t] % MOD ) % MOD;        }        if( sta[i - 1] != now[i - 1] ) diff--;    }    return res % MOD;}int main(){    scanf("%d %d" , &N , &K );    prework();    scanf("%s %s" , s , t );    ll res = solve( t , s );    cout << res % MOD << endl;    return 0;}

T2 Running

这道题感觉上是这套题的一个难点了，当时没有能够想出来，60分还是很好拿的，据说某些学佛写挂了，%%%。对于每一个人，显然他能够到达的格数是 gcd(Ai,N)。那么只需要枚举N的约数，每次加上ϕ(n/d)即为答案。

/*  ID:Agreement*/// Invincible#include <bits/stdc++.h>#define rep( i , l , r ) for( int i = (l) ; i <= (r) ; ++i )#define per( i , r , l ) for( int i = (r) ; i >= (l) ; --i )#define erep( i , u ) for( int i = head[(u)] ; ~i ; i = e[i].nxt )using namespace std;inline int _read(){    register int x = 0 , f = 1;    register char ch = getchar();    while( ch > '9' || ch < '0' ) { if( ch == '-' ) f = -1; ch = getchar(); }    while( ch >= '0' && ch <= '9' ){        x = x * 10 + ch - '0';        ch = getchar();    }    return x * f;}template<class T> inlinevoid read(T *a) {    char c;    while (isspace(c = getchar())) {}    bool flag = 0;    if (c == '-') flag = 1, *a = 0;    else        *a = c - 48;    while (isdigit(c = getchar())) *a = *a * 10 + c - 48;    if (flag) *a = -*a;}const int N = 10001;int n, m;int a[N], d[N], cnt;int phi(int x) {    int res = x;    REP(i, 1, cnt) if (x % d[i] == 0)        res = res / d[i] * (d[i] - 1);    return res;}int solve(int x) {    REP(i, 1, m) if (x % a[i] == 0) return phi(n / x);    return 0;}int main() {    cin >> n >> m;    REP(i, 1, m) read(a + i), a[i] = __gcd(a[i], n);    int t = n;    for (int i = 2; i * i <= n; ++i)         if (t % i == 0) {            d[++cnt] = i;            while (t % i == 0) t /= i;        }    if (t > 1) d[++cnt] = t;    int num = 0;    for (int i = 1; i * i <= n; ++i) if (n % i == 0) {        num += solve(i);        if (i * i != n) num += solve(n / i);    }    cout << n - num << endl;    return 0;}

T3 Tree

简单的树上背包，直接放代码了。

/*    ID:Agreement*///invincible#include <bits/stdc++.h>#define rep( i , l , r ) for( int i = (l) ; i <= (r) ; ++i )#define per( i , r , l ) for( int i = (r) ; i >= (l) ; --i )#define erep( i , u ) for( int i = head[(u)] ; ~i ; i = e[i].nxt )using namespace std;const int maxn = 3000 + 5;struct edge{    int v , nxt;} e[maxn * 2];int head[maxn] , _t = 0;inline void ins( int u , int v ){    e[_t].v = v , e[_t].nxt = head[u] , head[u] = _t; _t++;    e[_t].v = u , e[_t].nxt = head[v] , head[v] = _t; _t++;}const int INF = 0x3f3f3f3f;int N , lim , val[maxn];int f[maxn][maxn];void dfs( int d , int u , int fa , int c ){    erep( i , u ){        int v = e[i].v;        if( v == fa ) continue;        rep( t , 0 , c ) f[d + 1][t] = f[d][t];        dfs( d + 1 , v , u , c - 1 );        rep( t , 1 , c ) f[d][t] = max( f[d][t] , f[d + 1][t - 1] + val[v] );    }}int main(){    int u , v;    scanf("%d %d" , &N , &lim);    rep( i , 1 , N ) scanf("%d" , &val[i]);    memset( head , 0xff , sizeof head );    rep( i , 1 , N - 1 ){        scanf("%d %d" , &u , &v );        ins( u , v );    }    ins( 0 , 1 );    rep( i , 0 , lim ) f[1][i] = -INF;    f[1][0] = 0;    int ans = 0;    dfs( 1 , 0 , -1 , lim );    rep( i , 0 , lim ) ans = max( ans , f[1][i] );    cout << ans << endl;    return 0;}