181106 solution

今天心态爆炸

一直在练的树dp没做出来

以为自己没问题比较擅长的容斥也没做出来

最后其实T2自己的做法再想一想就是正解

算了说出来也没人信

T1如果沿用之前的思路应该也能想出来...

考试的时候不会再崩溃了

因为没资格

做题的顺序改一下吧 看啥做啥 觉得自己能做就想到底

不要以为自己能快速切某个题

如果真能快速想出来任何做法就写了

暴力以后对拍也能用上

这样想题才能专心

 

T1 matrix 20/100

没练容斥....

容斥就是多退少补

想了O(n^2)却没写

就是

(其实还是指+1以为是+i*j)

问题就在于如何化简

题解是先把n-i,n-j变成i,j

然后把j压到一起

把 -3ⁱ视为一项 然后...

这不就是个二项式定理吗

所以愉快解决

注意上面只能处理 i*j不为0的情况

为0暴力手算即可

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<iostream>
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
typedef double D;
#define eps 1e-8
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
//head
const int N = 1000006;
const ll mod = 998244353;
int n;
ll fac[N],inv[N];
ll ksm(ll a,ll b) {
    ll tmp = 1;
    while(b) {
        if(b & 1) tmp = tmp * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return tmp;
}

void initfac(int n) {
    fac[0] = fac[1] = inv[1] = inv[0] = 1;
    rep(i,2,n) fac[i] = fac[i-1] * i % mod;
    inv[n] = ksm(fac[n],mod - 2);
    per(i,n-1,2) inv[i] = inv[i+1] * (i+1) % mod;
}

ll C(ll n,ll m) {return fac[n] * inv[m] % mod * inv[n-m] % mod;}

ll ans = 0;
ll sgn(ll x) {return (x & 1) ? 1 : mod - 1;}
int main() {
    n = read();
    initfac(n);
    rep(i,1,n) ans = (ans + ksm(3,i) * ksm(3,n*(n-i)) % mod * 2 * sgn(i) % mod * C(n,i) % mod) % mod;
    rep(i,0,n-1) {
        ll tmp = 3ll * sgn(i) * C(n,i) % mod;
        ll tmpp = (mod - ksm(3,i)) % mod;
        tmpp = (ksm(tmpp + 1,n) - ksm(tmpp,n) + mod) % mod;
        tmp = tmp * tmpp % mod;
        ans = (ans + tmp) % mod;
    }
    printf("%lld\n",ans);
    return 0;
}

 

T2 Tree 0/100

树dp水题

并没有做出来

准确的说是做出来了并没有写

最后完全心态爆炸 如果敲一下怎么也有写跪那一帮人的30吧

先尽量减 判断是否合法

然后 重新 模拟减的过程

至于如何分配各个儿子减的值

选择在处理父亲的时候减掉儿子的边权

 

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<iostream>
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
typedef double D;
#define eps 1e-8
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
//head
const int N = 200006;
int n;
int head[N],nxt[N<<1],mo[N<<1],cnt;
ll cst[N<<1],maxx[N<<1];
void _add(int x,int y,ll A,ll B) {
    mo[++cnt] = y;
    nxt[cnt] = head[x];
    head[x] = cnt;
    cst[cnt] = A,maxx[cnt] = B;
}
void add(int x,int y) {
    ll A = read(),B = read();
    // printf("%d %d %d %d\n",x,y,A,B);
    _add(x,y,A,B),_add(y,x,A,B);
}

ll w[N],f[N];
void dfs(int x,int p) {
    for(int i = head[x];i;i = nxt[i]) {
        int sn = mo[i];
        if(sn == p) continue;
        dfs(sn,x);
        ll a = cst[i],b = maxx[i];
        w[x] += w[sn] + a;
        f[x] += f[sn] + a;
        ll C = min(a - 1,b - f[sn]);
        if(C < 0) puts("-1"),exit(0);
        f[x] -= C;
    }
}
ll dp[N],sum[N];
void Dp(int x,int p) {
    sum[x] = 0;
    for(int i = head[x];i;i = nxt[i]) {
        int sn = mo[i];
        if(sn == p) continue;
        Dp(sn,x);
        ll a = cst[i],b = maxx[i];
        if(b >= sum[sn]) {
            dp[x] += dp[sn],sum[x] += sum[sn] + a;
        } else {
            dp[x] += dp[sn] + sum[sn] - b;
            sum[sn] = b,sum[x] += a + b;
        }
    }
}

int main() {
    freopen("tree.in","r",stdin);
    freopen("tree.out","w",stdout);
    n = read();
    rep(i,2,n) add(read(),read());
    dfs(1,1),Dp(1,1);
    printf("%lld\n",dp[1]);
    return 0;
}

 

T3 count 10/100

dp 完全想不到维护多少条相交路径啊....

dp[i][j]表示i深度子树有j条不想交路径的方案数

所以可以根据路径是否经过根节点分类计数

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<iostream>
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
typedef double D;
#define eps 1e-8
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
//head
const int N = 405;
const int mod = 1e9+7;
int n;
ll fac[N],inv[N];
ll ksm(ll a,ll b) {
    ll tmp = 1;
    while(b) {
        if(b & 1) tmp = tmp * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return tmp;
}

void initfac(int n) {
    fac[0] = fac[1] = inv[1] = inv[0] = 1;
    rep(i,2,n) fac[i] = fac[i-1] * i % mod;
    inv[n] = ksm(fac[n],mod - 2);
    per(i,n-1,2) inv[i] = inv[i+1] * (i+1) % mod;
}

ll C(ll n,ll m) {return fac[n] * inv[m] % mod * inv[n-m] % mod;}

ll dp[N][N];
int main() {
    freopen("count.in","r",stdin);
    freopen("count.out","w",stdout);
    n = read();
    initfac(n);
    dp[1][0] = dp[1][1] = 1;
    rep(i,2,n) {
        rep(j,0,n) rep(k,0,n-j) {
            dp[i][j+k] = (dp[i][j+k] + dp[i-1][j] * dp[i-1][k] % mod) % mod;
            dp[i][j+k+1] = (dp[i][j+k+1] + dp[i-1][j] * dp[i-1][k] % mod) % mod;
            dp[i][j+k] = (dp[i][j+k] + (j+k<<1) * dp[i-1][j] % mod * dp[i-1][k] % mod) % mod;
            if(j+k) dp[i][j+k-1] = (dp[i][j+k-1] + (C(j+k,2)<<1) * dp[i-1][j] % mod * dp[i-1][k] % mod) % mod;
        }
    }
    printf("%lld\n",dp[n][1]);
    return 0;
}