文章目录

• AtCoder Beginner Contest 208
• • A - Rolling Dice
  • B - Factorial Yen Coin
  • C - Fair Candy Distribution
  • D - Shortest Path Queries 2
  • E - Digit Products
  • F - Cumulative Sum

AtCoder Beginner Contest 208

A - Rolling Dice

题意：

题解：

代码：

#include <bits/stdc++.h>

#define int long long
#define debug(x) cout << #x << " = " << x << endl;
using namespace std;

int const MAXN = 2e5 + 10;
int n, m, T, a[MAXN];
int A, B;

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> A >> B;
    int low = A;
    int up = A * 6;
    if (low <= B && B <= up) puts("Yes");
    else puts("No");
    return 0;
}

B - Factorial Yen Coin

题意：

题解：

代码：

#include <bits/stdc++.h>

#define int long long
#define debug(x) cout << #x << " = " << x << endl;
using namespace std;

int const MAXN = 2e5 + 10;
int n, m, T, a[MAXN];
int get(int x) {
    int res = 1;
    for (int i = 1; i <= x; ++i) res *= i;
    return res;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n;
    int cnt = 0;
    for (int i = 10; i >= 1; --i) {
        cnt += n / get(i);
        n %= get(i);
    }
    cout << cnt;
    return 0;
}

C - Fair Candy Distribution

题意：

题解：

代码：

#include <bits/stdc++.h>

#define int long long
#define debug(x) cout << #x << " = " << x << endl;
using namespace std;

int const MAXN = 2e5 + 10;
int n, m, T;
typedef pair<int, int> PII;
PII a[MAXN];

bool cmp(PII x, PII y) {
    return x.second < y.second;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i].first;
        a[i].second = i;
    }
    sort(a + 1, a + 1 + n);
    int avg = m / n;
    int remain = m % n;
    for (int i = 1; i <= n; ++i) {
        if (i <= remain) a[i].first = avg + 1;
        else a[i].first = avg;
    }
    sort(a + 1, a + 1 + n, cmp);
    for (int i = 1; i <= n; ++i) cout << a[i].first << endl;
    return 0;
}

D - Shortest Path Queries 2

题意： 有一张n个点m条边的有向图，定义 f ( s , t , k ) f(s,t,k) f(s,t,k) 为从s到t点，只在前1 ~ k个点之间流转的路径的最小长度。现在要你计算 ∑ s = 1 N ∑ t = 1 N ∑ k = 1 N f ( s , t , k ) \sum_{s=1}^N\sum_{t=1}^N\sum_{k=1}^Nf(s, t, k) ∑s=1N​∑t=1N​∑k=1N​f(s,t,k)。 数据范围 1 < = n < = 400 , 0 < = m < = n ∗ ( n − 1 ) 1<=n<=400,0<=m<=n*(n-1) 1<=n<=400,0<=m<=n∗(n−1)

题解： n是400，第一眼看过去应该就会想到Floyd。考虑floyd的过程，第一个循环for (int k = 1; k <= n; ++k)，这个循环决定了对于任意的i和j之间的转移，只经过当前的k来更新，因此当k从1 ~ n的过程，也就是使得i -> j之间的转移只通过1 ~ k的来流转的最短路，那么符合题目的定义。因此对于每一个k变化的时候，都更新下答案即可。

代码：

#include <bits/stdc++.h>

#define int long long
using namespace std;

int const N = 410, INF = 0x3f3f3f3f3f3f3f3f;
int mp[N][N], n, m, k;

// floyd求最短路
int floyd() {
    int res = 0 ;
    for (int k = 1; k <= n; ++k) {
        for (int i = 1; i <= n; ++i)  // 枚举起点
            for (int j = 1; j <= n; ++j)  // 枚举终点
                mp[i][j] = min(mp[i][j], mp[i][k] + mp[k][j]);  // 更新i到j的距离
        for (int i = 1; i <= n; ++i) 
            for (int j = 1; j <= n; ++j) 
                if (mp[i][j] < 1e9) res += mp[i][j];
    }
    return res;
}

signed main() {
    cin >> n >> m;  // 点、边、询问次数
    
    // 初始化距离：初始化为正无穷
    for (int i = 1; i <= n; ++i) 
        for (int j = 1; j <= n; ++j) 
            mp[i][j] = (i == j? 0: INF);  
            
    // 读入边
    for (int i = 1, a, b, c; i <= m; ++i) {
        scanf("%lld%lld%lld", &a, &b, &c);
        mp[a][b] = min(mp[a][b], c);
    }
    
    // 求最短路
    cout << floyd();
    
    return 0;
}

E - Digit Products

题意： 计算[1, N]中有多少数字的数位乘积小于等于K。 1 < = N < = 1 e 18 , 1 < = K < = 1 e 9 1<=N<=1e18,1<=K<=1e9 1<=N<=1e18,1<=K<=1e9

题解： 本题需要考虑乘积，有18位，乘积的数字很大，但是状态数目不会很多，因此直接把状态离散化。同时，需要考虑前导0的影响。然后直接按照数位dp转移即可。

代码：

#include <bits/stdc++.h>

#define int long long
#define debug(x) cout << #x << " = " << x << endl;
using namespace std;

int const MAXN = 20;
int n, k, a[MAXN], dp[MAXN][1000000], len, idx;
unordered_map<int, int> H;

int mapping(int x) {
    if (!H.count(x)) H[x] = ++idx;
    return H[x];
}   

int dfs(int pos, int lead, int sum, int limit) {  // lead维护前导0，sum维护乘积
    if (!pos) return sum <= k;
    if (!limit && !lead && dp[pos][mapping(sum)] != -1) return dp[pos][mapping(sum)];  // 因为乘积总数少，所以离散化
    int res = 0;
    int up = limit? a[pos]: 9;
    for (int i = 0; i <= up; ++i) {
        if (lead) res += dfs(pos - 1, lead && !i, max((int)1, i), limit && i == up);  // 需要考虑前导0带来的影响
        else res += dfs(pos - 1, lead && !i, sum * i, limit && i == up);
    }
    if (!limit && !lead) dp[pos][mapping(sum)] = res;
    return res;
}

int solve(int x) {
    len = 0;
    while(x) a[++len] = x % 10, x /= 10;
    return dfs(len, 1, 1, 1);
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n >> k;
    memset(dp, -1, sizeof dp);
    cout << solve(n) - 1;
    return 0;
}

F - Cumulative Sum

题意： 阴间拉格朗日插值，题解看了半天，还是不会，鸽了

题解：

代码：

标签：AtCoder,Beginner,int,208,sum,cin,long,题解,define
来源： https://blog.csdn.net/m0_49959202/article/details/119425203