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暑假集训五[星际旅行, 砍树, 超级树, 求和]

作者：互联网

暑假集训5

星际旅行

这个题刚看我觉得很ex，没事思路，就跳了，然后就去欺负\(T4\)了后来别的不会做，然后回来肝它...就肝出来了...对了，注意开\(long long\)

here 
#include <bits/stdc++.h>
#define LL long long
#define Re register int
#define LD long double
#define mes(x, y) memset(x, y, sizeof(x))
#define cpt(x, y) memcpy(x, y, sizeof(x))
#define fuc(x, y) inline x y
#define fr(x, y, z)for(Re x = y; x <= z; x ++)
#define fp(x, y, z)for(Re x = y; x >= z; x --)
#define frein(x) freopen(#x ".in", "r", stdin)
#define freout(x) freopen(#x ".out", "w", stdout)
#define ki putchar('\n')
#define fk putchar(' ')
#define WMX aiaiaiai~~
#define pr pair<long long, long long>
#define mk(x, y) make_posair(x, y)
using namespace std;
namespace kiritokazuto{
    auto read = [](){
        LL x = 0;
        int f = 1;
        char c;
        while (!isdigit(c = getchar())){ if (c == '-')f = -1; }
        do{ x = (x << 1) + (x << 3) + (c ^ 48); } while (isdigit(c = getchar()));
        return x * f;
    };
    template <typename T> fuc(void, write)(T x){
        if (x < 0)putchar('-'), x = -x;
        if (x > 9)write(x / 10); putchar(x % 10 | '0');
    }
}

using namespace kiritokazuto;
const int maxn = 1e5 + 10, Mod = 1e9 + 7;
const int Inf = 2147483647;
//无向图
//有自环
//话说这玩意真的不是排列组合吗?
//(23 24 25 34 35 45 。。。。。)

/*
本质不同
1> 起点不同
2> 终点不同
3> 走不同性质的边(点)
4> 当且仅当至少存在一个虫
洞，在两条航线中经过的次数不同
*/
/*
我行过许多地方的桥，看过许多次数的云，喝过许多不同种类的酒
却只爱过一个正当最好年纪的人
*/
//它相当于是让我走完，应该是一笔画
//话说昨天long long卡我那么多，今天全开long long 妈的
bool vis[maxn];
LL ans = 0;
LL deg[maxn];
int n;
int m;
struct Node{
    int to, next;
}wmx[maxn << 1];
int head[maxn], len = 0;
fuc(void, Qian)(int from, int to){
    wmx[++len].to = to;
    wmx[len].next = head[from];
    head[from] = len;
}
int cir, cnt;
//cir为自环的个数， cnt为正常边的个数
//又想暴力删边了
/*
！！！！
被删去的两条边一定连在同一个顶点上
否则删了他们之后就会有 4个度数为奇数的点
跑不了一笔画
连在一起则4 - 2 = 2还是偶数
然后我分情况讨论
1> 删两个自环
2> 删一个自环和一个边(不用在一个顶点上)
3> 删在同一个顶点上的两条边
*/
fuc(void, dfs_check)(int x){
    vis[x] = 1;
    for (Re i = head[x]; i; i = wmx[i].next){
        int to = wmx[i].to;
        if (!vis[to])dfs_check(to);
    }
}
signed main(){
    n = read();
    m = read();
    fr(i, 1, m){
        int x = read(), y = read();
        if (x == y){ cir++; continue; }//自环的deg不计入，不连边
        deg[x] ++;
        deg[y] ++;
        Qian(x, y);
        Qian(y, x);
    }
    cnt = m - cir;
    if (cir == m){
        printf("0");
        return 0;
        //全是自环啥也不行，图不联通
    }
    fr(i, 1, n){
        if (deg[i]){ dfs_check(i); break; }
    }//判断联通否
    fr(i, 1, n){
        if (deg[i] && !(vis[i])){//图不联通
            printf("0\n");
            return 0;
        }
    }
    fr(i, 1, n)ans += (deg[i] * (deg[i] - 1)) >> 1;//第三种
    ans += (cir * (cir - 1)) >> 1;//随便删两个自环，删完之后那两个点就成deg为奇的；了
    ans += cnt * cir;//两个组合
    write(ans);
    return 0;
}
/*
5 4
1 2
1 3
1 4
1 5

3 4
1 1
2 1
2 3
3 3
*/

砍树

这个题没有单调性，不能二分，呜呜呜

$ \begin{aligned}
&\frac{a}{b}=\left\lfloor\frac{a}{b}\right\rfloor+r(0\leq r<1)\
\implies
&\left\lfloor\frac{a}{bc}\right\rfloor
=\left\lfloor\frac{a}{b}\cdot\frac{1}{c}\right\rfloor
=\left\lfloor \frac{1}{c}\left(\left\lfloor\frac{a}{b}\right\rfloor+r\right)\right\rfloor
=\left\lfloor \frac{\left\lfloor\frac{a}{b}\right\rfloor}{c} +\frac{r}{c}\right\rfloor
=\left\lfloor \frac{\left\lfloor\frac{a}{b}\right\rfloor}{c}\right\rfloor\
&&\square
\end{aligned} $

fuc(bool, check)(int x){
    int res = 0;
    fr(i, 1, n){
        res += ceil(1.0 * Exp[i] / x);
    }
    return res * x <= C;
}
signed main(){
    n = read();
    k = read();
    fr(i, 1, n)Exp[i] = read(), C += Exp[i];
    C += k;
    for (Re l = 1, r; l <= C; l = r + 1){
        r = floor(C / (C / l));
        if (check(r)){
            ans = r;
        }
    }
    write(ans);
    return 0;
}

超级树

怎么天天都是神仙dp啊，放过我吧

here 
#include <bits/stdc++.h>
#define LL long long
#define Re register int
#define LD long double
#define mes(x, y) memset(x, y, sizeof(x))
#define cpt(x, y) memcpy(x, y, sizeof(x))
#define fuc(x, y) inline x y
#define fr(x, y, z)for(Re x = y; x <= z; x ++)
#define fp(x, y, z)for(Re x = y; x >= z; x --)
#define frein(x) freopen(#x ".in", "r", stdin)
#define freout(x) freopen(#x ".out", "w", stdout)
#define ki putchar('\n')
#define fk putchar(' ')
#define WMX aiaiaiai~~
#define pr pair<long long, long long>
#define mk(x, y) make_posair(x, y)
using namespace std;
namespace kiritokazuto{
    auto read = [](){
        LL x = 0;
        int f = 1;
        char c;
        while (!isdigit(c = getchar())){ if (c == '-')f = -1; }
        do{ x = (x << 1) + (x << 3) + (c ^ 48); } while (isdigit(c = getchar()));
        return x * f;
    };
    template <typename T> fuc(void, write)(T x){
        if (x < 0)putchar('-'), x = -x;
        if (x > 9)write(x / 10); putchar(x % 10 | '0');
    }
}

using namespace kiritokazuto;
const int maxn = 410, Mod = 1e9 + 7;
const int Inf = 2147483647;
#define int long long
int n;
int dp[1010][1010];
LL p = 0;
signed main(){
    n = read(), p = read();
    dp[1][0] = dp[1][1] = 1;
    fr(i, 1, n - 1){
        fr(l, 0, n){
            for (Re r = 0;l + r - 1 <= n; r++){
                LL num = dp[i][l] * dp[i][r] % p;
                dp[i + 1][l + r] = (dp[i + 1][l + r] + (2 * (l + r) % p + 1) % p * num) % p;
                dp[i + 1][l + r + 1] = (dp[i + 1][l + r + 1] + num) % p;
                if (l + r)dp[i + 1][l + r - 1] = (dp[i + 1][l + r - 1] + 2ll * num * l % p * r % p) % p, dp[i + 1][l + r - 1] = (dp[i + 1][l + r - 1] + num * (l * (l - 1) % p + r * (r - 1) % p) % p) % p;
            }
        }
    }
    write(dp[n][1] % p);
    return 0;
}

求和

签到题，没啥好说的...话说我暴力跑的好快啊，注释掉的是暴力，其实就是预处理一下，然后差分就行

here 
#include <bits/stdc++.h>
#define LL long long
#define Re register int
#define LD long double
#define mes(x, y) memset(x, y, sizeof(x))
#define cpt(x, y) memcpy(x, y, sizeof(x))
#define fuc(x, y) inline x y
#define fr(x, y, z)for(Re x = y; x <= z; x ++)
#define fp(x, y, z)for(Re x = y; x >= z; x --)
#define frein(x) freopen(#x ".in", "r", stdin)
#define freout(x) freopen(#x ".out", "w", stdout)
#define ki putchar('\n')
#define fk putchar(' ')
#define WMX aiaiaiai~~
#define pr pair<long long, long long>
#define mk(x, y) make_posair(x, y)
using namespace std;
namespace kiritokazuto{
    auto read = [](){
        LL x = 0;
        int f = 1;
        char c;
        while (!isdigit(c = getchar())){ if (c == '-')f = -1; }
        do{ x = (x << 1) + (x << 3) + (c ^ 48); } while (isdigit(c = getchar()));
        return x * f;
    };
    template <typename T> fuc(void, write)(T x){
        if (x < 0)putchar('-'), x = -x;
        if (x > 9)write(x / 10); putchar(x % 10 | '0');
    }
}

using namespace kiritokazuto;
const int maxn = 300100, Mod = 998244353;
const int Inf = 2147483647;
//虽然但是，D题看着确实好欺负
//先莽个暴力
//又额树剖干上了
struct Node{
    int to, next;
}wmx[maxn << 1];
LL head[maxn], len = 0;
fuc(void, Qian)(int from, int to){
    wmx[++len].to = to;
    wmx[len].next = head[from];
    head[from] = len;
}
int k, n, m;
LL ans;
LL top[maxn], sz[maxn], son[maxn], dep[maxn], fa[maxn];
LL Maxdep = -1;
fuc(void, dfs1) (int x, int pre){
    sz[x] = 1;
    if (x == 1)dep[x] = 0;
    else dep[x] = dep[pre] + 1;
    fa[x] = pre;
    Maxdep = max(Maxdep, dep[x]);
    for (Re i = head[x]; i; i = wmx[i].next){
        int to = wmx[i].to;
        if (to == pre)continue;
        dfs1(to, x);
        sz[x] += sz[to];
        if (sz[son[x]] < sz[to])son[x] = to;
    }
}
fuc(void, dfs2) (int x, int tp){
    top[x] = tp;
    if (!son[x])return;
    dfs2(son[x], tp);
    for (Re i = head[x]; i; i = wmx[i].next){
        int to = wmx[i].to;
        if (to != fa[x] && to != son[x])dfs2(to, to);
    }
}
fuc(int, get_lca)(int x, int y){
    if (top[x] == top[y]){
        return (dep[x] < dep[y]) ? x : y;
    }
    while (top[x] != top[y]){
        if (dep[top[x]] < dep[top[y]])swap(x, y);
        x = fa[top[x]];
    }
    // printf("x == %d y == %d\n", x, y);
    return (dep[x] < dep[y]) ? x : y;
}
LL Pow[maxn][51];
fuc(void, init)(LL x){
    fr(i, 1, x)Pow[i][1] = i;//所有dep自己的
    fr(i, 1, x){
        fr(j, 2, 50){
            Pow[i][j] = (Pow[i][j - 1] * i) % Mod;
            //累乘
        }
    }
    fr(j, 1, 50){
        fr(i, 1, x){
            Pow[i][j] = (Pow[i][j] + Pow[i - 1][j]) % Mod;
            //链前缀和
        }
    }
}
// fuc(LL, qpow)(LL a, LL b){
//     LL res = 1;
//     while (b){
//         if (b & 1) res = (res * a) % Mod;
//         a = (a * a) % Mod;
//         b >>= 1;
//     }
//     return res % Mod;

// }

//疏忽了，k <= 50打个表就行....qpow太慢了
//可以类比一下那个dis的操作

/*
这拍了一下，好像也没快多少....
1900 -> 1800
保佑能卡过去！
*/
signed main(){
    // frein(data);
    // freout(std);
    n = read();
    fr(i, 1, n - 1){
        int x = read(), y = read();
        Qian(x, y);
        Qian(y, x);
    }
    dfs1(1, 0);
    dfs2(1, 1);
    //ri dfs2写成dfs1了,浪费半个小时
    m = read();
    // fr(i, 1, n){
    //     printf("dep[%d] = %d  top[%d] = %d\n", i, dep[i], i, top[i]);
    // }
    init(Maxdep);
    fr(i, 1, m){
        ans = 0;
        int x = read(), y = read(), k = read();
        int lca = get_lca(x, y);
        ans = Pow[dep[x]][k] + Pow[dep[y]][k] - Pow[dep[lca]][k];
        if (dep[lca] > 1)ans -= Pow[dep[lca] - 1][k];
        ans = (ans % Mod + Mod) % Mod;
        // ans = (ans + qpow(dep[lca], k)) % Mod;
        // int st = 0;
        // st = (dep[x] < dep[y]) ? x : y;
        // int ed = 0;
        // ed = (st == x) ? y : x;
        // // printf("lca = %d st = %d ed = %d ans = %lld dep[%d] = %d\n", lca, st, ed, ans, lca, dep[lca]);
        // if (top[x] != top[y])fr(j, dep[lca] + 1, dep[st]) ans = (ans + 2 * qpow(j, k)) % Mod;
        // else fr(j, dep[lca] + 1, dep[st]) ans = (ans + qpow(j, k)) % Mod;
        // // printf("now ans = %lld \n", ans);
        // fr(j, dep[st] + 1, dep[ed]) ans = (ans + qpow(j, k)) % Mod;

        write(ans);
        ki;
    }
    return 0;
}
/*
5
1 2
1 3
2 4
2 5
2
1 4 5
5 4 45
*/

标签：fr,int,long,dep,砍树,暑假,ans,集训,define
来源： https://www.cnblogs.com/kiritokazuto/p/16596971.html