Summary

这场比赛本来是一次上分的好机会，一开始做出来了四道题，Rank排在前300，结果B因为看错题并且PP太水而惨遭叉掉，Rank直接掉到1600+，又要掉分了 以后打CF一定一定一定一定一定一定一定一定在做完会做的题之后要检查之前的每一道题，每一回打CF都是想当然觉得应该没问题就没管了，导致惨遭FST

Problems

A

绝对值相加即可

B

按照题意模拟，把减法改成取模即可

C

比较简单的模拟（贪心），对于每个字母它能变成排在它之后的所有字母，然后依次贪心即可 

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#include <bits/stdc++.h>

using namespace std;

const int Maxn = 1e5 + 10;

char S[Maxn];

int N, M;
int A[Maxn];

int main()
{
#ifdef hk_cnyali
    freopen("C.in", "r", stdin);
    freopen("C.out", "w", stdout);
#endif
    scanf("%s", S + 1);
    int N = strlen(S + 1);
    if (N < 26)
    {
        cout<<-1<<endl;
        return 0;
    }
    int pos = 1;
    for (int i = 1; i <= N; ++i)
    {
        if (S[i] - 'a' + 1 <= pos && pos <= 26)
        {
            S[i] = 'a' + pos - 1;
            ++pos;
        }
    }
    if (pos <= 26) cout<<-1<<endl;
    else puts(S + 1);
    return 0;
}

D

设

TeX parse error: Undefined control sequence \[

表示第i天逃j次课最少需要花费的逃课次数，

TeX parse error: Undefined control sequence \[

表示到第i天总共逃了j次课最少需要花费的逃课次数。对于WW数组可以通过前缀和N^3方计算出来，Dp数组也是一个比较水的Dp 
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#include <bits/stdc++.h>

using namespace std;

const int Maxn = 500 + 10, inf = INT_MAX;

char S[Maxn];

int N, M, K, A[Maxn][Maxn];
int w[Maxn][Maxn];
int Sum[Maxn][Maxn];
int Dp[Maxn][Maxn];

int main()
{
#ifdef hk_cnyali
    freopen("D.in", "r", stdin);
    freopen("D.out", "w", stdout);
#endif
    scanf("%d%d%d", &N, &M, &K);
    for (int i = 1; i <= N; ++i)
    {
        scanf("%s", S);
        for (int j = 1; j <= M; ++j)
            A[i][j] = S[j - 1] - '0', Sum[i][j] = Sum[i][j - 1] + A[i][j];
    }

    memset(w, 0x3f, sizeof (w));
    memset(Dp, 0x3f, sizeof (Dp));
    for (int i = 1; i <= N; ++i)
    {
        w[i][Sum[i][M]] = 0;
        for (int x = 1; x <= M; ++x)
            for (int y = x; y <= M; ++y)
            {
                int sum = Sum[i][M] - Sum[i][y] + Sum[i][x - 1];
                w[i][sum] = min(w[i][sum], y - x + 1);
            }
    }

    Dp[0][0] = 0;
    for (int i = 1; i <= N; ++i)
    {
        for (int j = 0; j <= K; ++j)
            for (int k = 0; k <= j; ++k)
                Dp[i][j] = min(Dp[i][j], Dp[i - 1][j - k] + w[i][k]);
    }
    int ans = inf;
    for (int i = 0; i <= K; ++i)
        ans = min(ans, Dp[N][i]);
    cout<<ans<<endl;
    return 0;
}