11.1NOIP模擬賽解題報告

自為風月馬前卒發表於2018-11-01

心路歷程

預計得分：(100 + 100 + 50)

實際得分：(100 + 100 + 50)

感覺老師找的題有點水呀。

上來看T1，woc？裸的等比數列求和？然而我不會公式呀。。感覺要涼

T2應該比較簡單，T3 dp能拿很多部分分。

但是T1只打暴力感覺好丟人啊。。想了10min發現不用公式也能做，就直接倍增一下就好了。

T2水題。感覺比T1還簡單。。

T3。。。。。這個就比較厲害了呀。賽後我大概問了一下，發現全機房一共讀出了(4)種題意Orzzz。

然後我花了(2h)做了一道水題。。然後發現錯誤的時候考試馬上就結束了，然後只能打個暴力走人。。。

T1

Orz zbq現場推出等比數列求和公式

Orz 好像除了我都會等比數列求和公式

Orzzzzzzzzzzzzzzzzz

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
#include<cmath>
#include<iostream>
using namespace std;
const int MAXN =1e5 + 10, mod = 1e9 + 7;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < `0` || c > `9`) {if(c == `-`) f = -1; c = getchar();}
    while(c >= `0` && c <= `9`) x = x * 10 +  c - `0`, c = getchar();
    return x * f;
}
int add(int x, int y) {
    if(x + y < 0) return x + y + mod;
    return x + y >= mod ? x + y - mod : x + y;
}
int mul(int x, int y) {
    return 1ll * x * y % mod;
}
int fp(int a, int p) {
    int base = 1;
    while(p) {
        if(p & 1) base = mul(base, a);
        a = mul(a, a); p >>= 1;
    }
    return base;
}
int N, M, pok[MAXN], g[MAXN];
int solve(int k, int n) {
    int len = 1;
    while((1ll << len) <= n) len <<= 1;
    pok[0] = k;
    for(int i = 1; i <= len; i++) pok[i] = mul(pok[i - 1], pok[i - 1]);
    g[0] = k;
    for(int i = 1; i <= len; i++) g[i] = add(g[i - 1], mul(g[i - 1], pok[i - 1]));
    int ans = 0, now = 0, base = 1;
    for(int i = len; i >= 0; i--) 
        if(now + (1 << i) <= n) 
            ans = add(ans, mul(g[i], base)), base = mul(base, pok[i]), now += (1 << i);
    return ans;
}
main() {
    freopen("sum.in", "r", stdin);
    freopen("sum.out", "w", stdout);
    N = read(); M = read();
    int ans = 0;
    for(int i = 1; i <= N; i++) {
        if(M & 1) ans = add(ans, add(solve(i, M - 1), fp(i, M)));
        else ans = add(ans, solve(i, M)); 
    //  cout << ans << endl;
    }
    cout << ans;
    return 0;
}

T2

(ans = all – min(sum[i]))

all表示所有邊權和

(sum[i])表示第(i)個節點到根的路徑

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
#include<cmath>
#include<iostream>
#define Pair pair<int, int>
#define MP make_pair
#define fi first
#define se second 
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9 + 7;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < `0` || c > `9`) {if(c == `-`) f = -1; c = getchar();}
    while(c >= `0` && c <= `9`) x = x * 10 +  c - `0`, c = getchar();
    return x * f;
}
int N, sum[MAXN], All;
vector<Pair> v[MAXN];
void dfs(int x, int fa) {
    for(int i = 0, to; i < v[x].size(); i++) {
        if((to = v[x][i].fi) == fa) continue;
        sum[to] = sum[x] + v[x][i].se;
        dfs(to, x);
    }
}
int main() {
    freopen("tour.in", "r", stdin);
    freopen("tour.out", "w", stdout);
    N = read();
    for(int i = 1; i <= N - 1; i++) {
        int x = read(), y = read(), z = read(); All += z; 
        v[x].push_back(MP(y, z));
        v[y].push_back(MP(x, z));
    }
    dfs(1, 0);
    All <<= 1;
    int ans = INF;
    for(int i = 1; i <= N; i++) ans = min(ans, All - sum[i]);
    cout << ans;
    return 0;
}

T3

神仙閱讀理解題，不過還是挺interesting的

首先，序列內的元素是無序的，這樣我們可以對相同的數字一起考慮

稍微想一下不難發現，幸運數字最多有(2^9)個

直接(f[i][j])表示前(i)個數，選(j)的方案，dp一下

最後合併答案的時候揹包一下

#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<set>
#include<bitset>
#include<iostream>
#include<map>
#define Pair pair<int, int>
#define MP make_pair
#define fi first
#define se second 
//#define int long long 
using namespace std;
const int MAXN = 1e5 + 10, mod = 1e9 + 7;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < `0` || c > `9`) {if(c == `-`) f = -1; c = getchar();}
    while(c >= `0` && c <= `9`) x = x * 10 + c - `0`, c = getchar();
    return x * f;
}
int N, K, a[MAXN], tot, cnt, fac[MAXN], ifac[MAXN];
map<int, int> mp;
int add(int &x, int y) {
    if(x + y < 0) x = x + y + mod;
    else x = (x + y >= mod ? x + y - mod : x + y);
}
int add2(int x, int y) {
    if(x + y < 0) return x + y + mod;
    else return x + y >= mod ? x + y - mod : x + y;
}
int mul(int x, int y) {
    return 1ll * x * y % mod;
}
int fp(int a, int p) {
    int base = 1;
    while(p) {
        if(p & 1) base = mul(base, a);
        a = mul(a, a); p >>= 1;
    }
    return base;
}
int C(int N, int M) {
    if(N < M) return 0;
    else return mul(fac[N], mul(ifac[M], ifac[N - M]));
}
int get(int x) {
    while(x) {
        if(x % 10 != 4 && x % 10 != 7) return 0;
        else x /= 10;
    }
    return 1;
}
map<int, Pair> id;
int rev[MAXN], f[2333][2333];
signed main() {
    freopen("lucky.in", "r", stdin);
    freopen("lucky.out", "w", stdout);
    N = read(); K = read();
    fac[0] = 1;
    for(int i = 1; i <= N; i++) fac[i] = mul(i, fac[i - 1]);
    ifac[N] = fp(fac[N], mod - 2);
    for(int i = N; i >= 1; i--) ifac[i - 1] = mul(ifac[i], i);
    
    for(int i = 1; i <= N; i++) {
        a[i] = read();
        if(get(a[i])) {
            if(!id[a[i]].fi) id[a[i]].fi = ++cnt, rev[cnt] = a[i];
            id[a[i]].se++;
        } else tot++;
    }
    f[0][0] = 1;
    for(int i = 1; i <= cnt; i++) {
        f[i][0] = 1;
        for(int j = 1; j <= cnt; j++) 
            f[i][j] = add2(f[i - 1][j], mul(f[i - 1][j - 1], id[rev[i]].se));       
    } 

    int ans = 0;
    for(int i = 0; i <= cnt; i++) add(ans, mul(f[cnt][i], C(tot, K - i)));
    cout << ans;
    
    return 0;
}

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