20200314题解

CCF T1

分析只有当n=1或n=2或n=5时才会买不完，除此之外尽量买一套，剩余情况直接暴力枚举吧

#include <iostream>
#include <cstdio>
#include <cstring>
#define LD long double
using namespace std;

long long read(){
    long long ans = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
        f *= (ch == '-') ? -1 : 1, ch = getchar();
    do ans = ((ans << 1) + (ans << 3) + (ch ^ 48)), ch = getchar();
    while(ch >= '0' && ch <= '9');
    return ans * f;
}

int main(){
    int n = read(), a = 0, b = 0, c = 0;
    a = b = c = n / 14, n %= 14;
    if(n == 1 || n == 2 || n == 5){
        if(!a){
            printf("-1");
            return 0;
        }
        n += 14, a--, b--, c--;
    }
    int a2 = -a-1, b2 = -b-1, c2 = -c-1;
    for(int i=0; i<=n; i+=7)
        for(int j=0; i+j<=n; j+=4)
            for(int k=0; i+j+k<=n; k+=3)
                if(i + j + k == n && i/7+j/4+k/3 > a2+b2+c2)
                    a2 = i / 7, b2 = j / 4, c2 = k / 3;
    printf("%d %d %d", a+a2, b+b2, c+c2);
    return 0;
}

CCF T2

裸的五边形数定理应用，打出模板就完了

#include <iostream>
#include <cstdio>
#include <cstring>
#define LD long double
using namespace std;

long long read(){
    long long ans = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
        f *= (ch == '-') ? -1 : 1, ch = getchar();
    do ans = ((ans << 1) + (ans << 3) + (ch ^ 48)), ch = getchar();
    while(ch >= '0' && ch <= '9');
    return ans * f;
}

const int MAXN = 100005;
long long f[MAXN], Dp[MAXN];

int main(){
    int n = read(), MOD = read();
    for(int i=1; f[f[0]] <= n; i++){
        f[++f[0]] = i * (3 * i - 1) / 2;
        f[++f[0]] = i * (3 * i + 1) / 2;
    }
    Dp[0] = 1;
    for(int i=1; i<=n; i++){
        for(int j=1; j<=f[0] && f[j]<=i; j++)
            Dp[i] = (Dp[i] + MOD + Dp[i-f[j]] * (((j - 1) % 4 <= 1) ? 1 : -1)) % MOD;
    }
    printf("%d", Dp[n]);
    return 0;
}

CCF T3

矩阵乘法，第a个矩阵val[i][j]表示从i到j用a次魔法的花费，快速幂即可。第1个矩阵可以通过枚举边O(n^2m)时间复杂度初始化

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

long long read(){
    long long ans = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
        f *= (ch == '-') ? -1 : 1, ch = getchar();
    do ans = ((ans << 1) + (ans << 3) + (ch ^ 48)), ch = getchar();
    while(ch >= '0' && ch <= '9');
    return ans * f;
}

const int MAXN = 101, MAXM = 2501;
struct Matrix{
    int size;
    long long value[MAXN][MAXN];
    void normal(){
        for(int i=1; i<=size; i++)
            for(int j=1; j<=size; j++)
                value[i][j] = (i == j) ? 0 : 1e15;
    }
    void operator *= (Matrix &other){
        long long ans[MAXN][MAXN];
        memset(ans, 127, sizeof(ans));
        for(int i=1; i<=size; i++)
            for(int j=1; j<=size; j++)
                for(int k=1; k<=size; k++)
                    ans[i][j] = min(ans[i][j], value[i][k] + other.value[k][j]);
        for(int i=1; i<=size; i++)
            for(int j=1; j<=size; j++)
                value[i][j] = ans[i][j];
    }
    void operator ^= (long long n){
        Matrix unit = (*this);
        this->normal();
        while(n){
            if(n & 1) (*this) *= unit;
            unit *= unit, n >>= 1;
        }
    }
}g;

int U[MAXM], V[MAXM], C[MAXM];
long long Dp[MAXN][MAXN];

int main(){
    int N = read(), M = read(), K = read();
    g.size = N;
    for(int i=1; i<=N; i++)
        for(int j=1; j<=N; j++)
            Dp[i][j] = g.value[i][j] = (i == j) ? 0 : 1e15;
    for(int i=1; i<=M; i++){
        U[i] = read(), V[i] = read(), C[i] = read();
        Dp[U[i]][V[i]] = C[i];
    }
    for(int k=1; k<=N; k++)
        for(int i=1; i<=N; i++)
            for(int j=1; j<=N; j++)
                Dp[i][j] = min(Dp[i][j], Dp[i][k] + Dp[k][j]);
    if(!K){
        printf("%lld", Dp[1][N]);
        return 0;
    }
    for(int i=1; i<=N; i++)
        for(int j=1; j<=N; j++)
            for(int k=1; k<=M; k++)
                g.value[i][j] = min(g.value[i][j], Dp[i][U[k]] + Dp[V[k]][j] - C[k]);
    g ^= K;
    printf("%lld", g.value[1][N]);
    return 0;
}

test

树链剖分，主要难点在于op=3时的操作，设dfs时根为1，当前根为root，求的是x的子树，分为3种情况

1.x=root，即求root的子树即可

2.x在root到1的路径上上，这种最麻烦，需要用整个树的权减去x的儿子中朝向root那个儿子son的树的权，son可以通过root来找

3.其它情况就求x的子树即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int read(){
    int ans = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
        f *= (ch == '-') ? -1 : 1, ch = getchar();
    do ans = ((ans << 1) + (ans << 3) + (ch ^ 48)), ch = getchar();
    while(ch >= '0' && ch <= '9');
    return ans * f;
}

const int MAXN = 100005, LOGN = 17;
struct G{
    int head[MAXN], last[MAXN<<1], next[MAXN<<1], lineNum;
    void add(int x,int y){
        lineNum++, next[lineNum] = head[x], last[lineNum] = y, head[x] = lineNum;
    }
}g;

int val[MAXN], dep[MAXN], size[MAXN], son[MAXN], fa[MAXN][LOGN+1], id[MAXN], top[MAXN];

struct BIT{
    int val[MAXN];
    void add(int pos,int x){
        for(int i=pos; i<MAXN; i+=i&(-i))
            val[i] += x;
    }
    int query(int pos){
        int ans = 0;
        for(int i=pos; i; i-=i&(-i))
            ans += val[i];
        return ans;
    }
    inline int query(int l,int r){
        return query(r) - query(l - 1);
    }
}bit;

void dfs1(int x,int faNum){
    dep[x] = dep[faNum] + 1, size[x] = 1, fa[x][0] = faNum;
    for(int i=1; i<=LOGN; i++)
        fa[x][i] = fa[fa[x][i-1]][i-1];
    for(int l=g.head[x]; l; l=g.next[l]){
        int y = g.last[l];
        if(y == faNum) continue;
        dfs1(y, x), size[x] += size[y];
        if(size[y] > size[son[x]]) son[x] = y;
    }
}

void dfs2(int x,int topNum){
    id[x] = ++id[0], top[x] = topNum;
    if(son[x]) dfs2(son[x], topNum);
    for(int l=g.head[x]; l; l=g.next[l]){
        int y = g.last[l];
        if(y != fa[x][0] && y != son[x])
            dfs2(y, y);
    }
}

int sum(int x,int y){
    int ans = 0;
    while(top[x] != top[y]){
        if(dep[top[x]] < dep[top[y]]) swap(x, y);
        ans += bit.query(id[top[x]], id[x]);
        x = fa[top[x]][0];
    }
    if(dep[x] > dep[y]) swap(x, y);
    ans += bit.query(id[x], id[y]);
    return ans;
}

inline int LCA(int x,int y){
    if(dep[x] < dep[y]) swap(x, y);
    for(int i=LOGN; i>=0; i--)
        if(dep[fa[x][i]] >= dep[y])
            x = fa[x][i];
    if(x == y) return x;
    for(int i=LOGN; i>=0; i--)
        if(fa[x][i] != fa[y][i])
            x = fa[x][i], y = fa[y][i];
    return fa[x][0];
}

inline int LCA(int x,int y,int root){
    int lx = LCA(x, root), ly = LCA(y, root);
    if(lx == ly) return LCA(x, y);
    return (dep[lx] > dep[ly]) ? lx : ly;
}

int main(){
    int N = read(), Q = read();
    for(int i=1; i<N; i++){
        int x = read(), y = read();
        g.add(x, y), g.add(y, x);
    }
    for(int i=1; i<=N; i++)
        val[i] = read();
    dfs1(1, 0);
    dfs2(1, 1);
    for(int i=1; i<=N; i++)
        bit.add(id[i], val[i]); 
    int root = 1;
    for(int i=1; i<=Q; i++){
        int op = read(), x = read();
        if(op == 1) root = x;
        else if(op == 2){
            int v = read();
            bit.add(id[x], v - val[x]);
            val[x] = v;
        }else if(op == 3){
            if(x == root) printf("%d\n", bit.query(1, N));
            else if(id[root] <= id[x] && id[x] <= id[root] + size[root] - 1) printf("%d\n", bit.query(id[x], id[x] + size[x] - 1));
            else if(id[x] <= id[root] && id[root] <= id[x] + size[x] - 1){
                int temp = root;
                for(int i=LOGN; i>=0; i--)
                    if(dep[fa[temp][i]] > dep[x])
                        temp = fa[temp][i];
                printf("%d\n", bit.query(1, N) - bit.query(id[temp], id[temp] + size[temp] - 1));
            }else printf("%d\n", bit.query(id[x], id[x] + size[x] - 1));
        }else if(op == 4){
            int y = read();
            int lca = LCA(x, y, root);
            printf("%d\n", sum(x, lca) + sum(y, lca) - val[lca]);
        }
    }
    return 0;
}