[C++知识库]AtCoder Beginner Contest 242

Problem A

• 求获取T恤的概率
• 分类讨论

#include <iostream>
#include <string>
#include <cstdio>
#include <limits.h>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <utility>
#include <unordered_set>
#include <queue>
#include <deque>
using namespace std;
const int N = 1e3+10;
const int M = 1e6;

int main() {
    int A, B, C, X;
    cin >> A>> B >> C >> X;
    if(X <= A) cout<<1<<endl;
    else if(X > B) cout<<0<<endl;
    else{
        printf("%.7lf", C*1.0/(B - A));
    }
    return 0;
}

Problem B

• 对字符串排序，从小到大输出

#include <iostream>
#include <string>
#include <cstdio>
#include <limits.h>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <utility>
#include <unordered_set>
#include <queue>
#include <deque>
using namespace std;
#include <iostream>
#include <string>
#include <cstdio>
#include <limits.h>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <utility>
#include <unordered_set>
#include <queue>
#include <deque>
using namespace std;
const int N = 1e3+10;
const int M = 1e6;

int main() {
    string s;
    cin >>s;
    sort(s.begin(), s.end());
    cout<<s<<endl;
    return 0;
}

Problem C

• 二维动态规划
• 初始的数值只能是1到9，注意边界数值1和9的转移方式。
• 注意数据范围是long long

#include <iostream>
#include <string>
#include <cstdio>
#include <limits.h>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <utility>
#include <unordered_set>
#include <queue>
#include <deque>
using namespace std;
const int N = 1e3+10;
const int M = 1e6;

constexpr int mod = 998244353;
LL dp[1000010][12];
LL cnt;

int main() {

    cin >> n;
    memset(dp, 0, sizeof dp);
    for(int i = 1; i <= 9; i++) dp[0][i] = 1;
    for(int i = 1; i < n; i++) {
        for(int j = 1; j <= 9; j++) {
            if(j == 1 ) dp[i][j] = (dp[i-1][j] + dp[i-1][j+1]) % mod;
            else if(j == 9)  dp[i][j] = (dp[i-1][j] + dp[i-1][j-1]) % mod;
            else{
                dp[i][j] = (dp[i-1][j-1] + dp[i-1][j] + dp[i-1][j+1]) % mod;
            }
            if(i == n-1) cnt = (cnt + dp[i][j]) % mod;
        }

    }
    cout<<cnt<<endl;
    return 0;
}