暑假第十四测

题解：第一题：

dp[i] 表示我们给这棵树再分配了i的度数

dp[i] = max(dp[j] + f[i - j + 1] - f[1]), 因为我们只考虑连在叶子节点上，所以只有叶子节点贡献改变

 

#include<bits/stdc++.h>
using namespace std;
const int M = 2020;
int n, val[M], dp[M];

int main(){
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        for(int i = 1; i < n; i++)scanf("%d", &val[i]);
        memset(dp, 0, sizeof(dp));
        dp[0] = n * val[1];
        for(int i = 1; i < n - 1; i++)
            for(int j = i - 1; j >= 0; j--)
                dp[i] = max(dp[i], dp[j] + val[i - j + 1] - val[1]);
        if(n == 1)printf("0
");
        else printf("%d
", dp[n - 2]);
    }
    
}

第二题：首先去除这个很烦的mod : ∑(i = 1~n) (k - k / i * i )　－－＞　　n*k -  ∑(i = 1~(min(n, k)) (k / i * i ),

k/i 有很多重复，分块求等差数列

#include<bits/stdc++.h>
using namespace std;
#define ll unsigned long long
ll ans;
int main(){
    freopen("b.in","r",stdin);
    freopen("b.out","w",stdout);
    ll n, k;
    scanf("%I64d%I64d", &n, &k);
        
    ans += n * k;
    ll res = min(n, k);
    ll lf = 1, rg;
    while(lf <= res){
        rg = k / (k/lf);
        if(rg > res) rg = res;
        ans -= (lf + rg) * (rg - lf + 1) / 2 * (k / lf);
        lf = rg + 1;
    }
    printf("%I64d
", ans);    

    
}

第三题：

#include<bits/stdc++.h>
using namespace std;
int n, m, k;

bool check(int p){
    long long ret = 0;
    if(k - 1 >= p - 1){
        ret += 1LL * p * (p - 1) / 2;
        ret += 1LL *(k - p);
    }
    else ret += 1LL*(2 * p - k) * (k - 1) / 2;
    
    if(n - k >= p - 1){
        ret += 1LL * p * (p - 1) / 2;
        ret += 1LL * (n - k - p + 1);
    } 
    else ret += 1LL * (2*p - 1 - n + k) * (n - k) / 2;
    return ret <= m - p;
    
}


int main(){
    freopen("c.in","r",stdin);
    freopen("c.out","w",stdout);
    int T;
    scanf("%d", &T);
    while(T--){
        int ans = 1;
        scanf("%d%d%d", &n, &m, &k);
        int rg = m - n + 1, lf = 1;
        while(lf <= rg){
            int mid = (lf + rg) >> 1;
            if(check(mid))ans = mid, lf = mid + 1;
            else rg = mid - 1;
        }
        printf("%d
", ans);
    }
}