A - Find Takahashi

找到序列中最高的数存在的位置

#include<bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int32_t main() {
    int n = read() , res = INT_MIN , id = 0;
    for( int i = 1 , x ; i <= n ; i ++ ){
        x = read();
        if( x > res ) res = x , id = i;
    }
    cout << id << "\n";

    return 0;
}

B - ABC-DEF

注意乘法可能会爆精度，所以注意取模

#include<bits/stdc++.h>
#define int long long
using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

const int mod = 998244353;

int32_t main() {
    int a = read() % mod , b = read() % mod , c = read() % mod , d = read() % mod ;
    int e = read() % mod , f = read() % mod;
    a = a * b % mod * c % mod;
    d = d * e % mod * f % mod;
    cout << ( ( a - d ) % mod + mod ) % mod;
    return 0;
}

C - Counting Squares

#代表棋子，问由棋子做顶点可以组合出多少个正方形。

首先先用 dfs 枚举出四个棋子。如果四个点两两之间的距离只有两种情况，那么该矩形就是一个正方形

#include<bits/stdc++.h>
#define int long long
using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

const int mod = 998244353;
int res;
vector< pair<int,int> > p , t;

int  dis( int ax , int ay , int bx , int by ){
    return ( ax - bx ) * ( ax - bx ) + ( ay - by ) * ( ay - by );
}
void dfs( int x ){
    if( t.size() == 4 ){
        set<int> s;
        for( int i = 0 ; i < 4 ; i ++ )
            for( int j = 0 ; j < i ; j ++ )
                s.insert( dis( t[i].first , t[i].second , t[j].first , t[j].second ) );
        if( s.size() == 2 ) res ++;
        return;
    }
    for( int i = x ; i < p.size() ; i ++ )
        t.push_back( p[i] ) , dfs( i+1 ) , t.pop_back();
    return;
}

int32_t main() {
    string s[10];
    for( int i = 0 ; i < 9 ; i ++ )
        cin >> s[i];
    for( int i = 0 ; i < 9 ; i ++ )
        for( int j = 0 ; j < 9 ; j ++ )
            if( s[i][j] == '#' ) p.push_back( { i , j } );
    dfs( 0 );
    cout << res << "\n";
    return 0;
}

D - Yet Another Recursive Function

这道题看似数据范围很大，但是因为是除法的问题，所以一共不会使用超过 100 个数组。发现这点的话就是一个简单的记忆化搜索，记忆化用个 map 存一下就好

#include<bits/stdc++.h>
#define int long long
using namespace std;

map<int,int> f;

int F( int x ){
    if( f.count(x) ) return f[x];
    return f[x] = F( x / 2 ) + F( x / 3 );
}
int32_t main() {
    int n;
    cin >> n;
    f[0] = 1;
    cout << F(n) << "\n";
	return 0;
}