codeforces1152 div2

比赛的链接

C

gcd(a+k, b+k) == gcd(a+k, b-a)

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
typedef long long ll;
ll ans=0;
ll mn;
ll a, b;

void work(ll x){
    ll k = (x-a%x)%x;
    ll aa = k+a, bb = k+b;
    ll temp = aa/__gcd(aa, bb)*bb;
    if(temp<mn){
        ans = k;
        mn = temp;
    }
    else if(temp == mn &&ans>k){
        ans = k;
    }
}

int main(){
    scanf("%lld%lld", &a, &b);
    if(a>b) swap(a, b);
    if(a == b){
        printf("0\n");
        return 0;
    }
    mn = a/__gcd(a, b)*b;
    ll d = b-a;
    for(int i=1; i*i<=d; ++i){
        if(d%i==0){
            work(1ll*i), work(1ll*d/i);
        }
    }
    printf("%lld\n", ans);

    return 0;
}

D

题意

括号串形成的trie树的最大的匹配（选择的两条边不能有公共的节点）。

记忆化dp

奇数层必定有孩子，因此只能取得一个边

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e3+10;
typedef long long ll;
const int mod = 1e9+7;
int dp[maxn][maxn];
//奇数层一定有孩子
int dfs(int n, int now){
    if(n == 0){
        if(now == 0) return dp[n][now] = 0;
        else return dp[n][now] = -2;
    }
    if(dp[n][now]!=-1) return dp[n][now];
    if(n<now||now<0) return dp[n][now] = -2;
    ll temp = 0;
    bool has = false;
    if(dfs(n-1, now+1)>=0){
        temp += dp[n-1][now+1]+(n%2==0);
        has = true;
    }
    if(dfs(n-1, now-1)>=0){
        temp += dp[n-1][now-1]+(n%2==0);
        has = true;
    }
    if(has){
        return dp[n][now] = temp%mod;
    }
    else return dp[n][now] = -2;
}

int main(){
    int n;
    scanf("%d", &n);
    memset(dp, -1, sizeof(dp));
    dfs(2*n, 0);
    printf("%d\n", dp[2*n][0]);
    return 0;
}