长乐Day 5

T1 又是一道暴力，直接预处理再枚举即可 AC代码

#include <bits/stdc++.h>
using namespace std;
int n;
string s1[4], s2[4], s3[4][7], s4[4][7];  // s2[i]是反转  s3[i]后k个  s[i] 前k个
int main() {
    freopen("restore.in", "r", stdin);
    freopen("restore.out", "w", stdout);
    cin >> n;
    cin >> s1[1] >> s1[2] >> s1[3];
    for (int i = 1; i <= 3; i++) {
        for (int j = n - 1; j >= 0; j--) s2[i] += s1[i][j];
    }
    for (int i = 1; i <= 3; i++) {
        for (int j = 0; j < n; j++) {
            for (int k = 0; k <= 6; k++) {
                if (s1[i][j] + k > 122)
                    s3[i][k] += (s1[i][j] + k - 26);
                else
                    s3[i][k] += (s1[i][j] + k);
            }
        }
    }
    for (int i = 1; i <= 3; i++) {
        for (int j = 0; j < n; j++) {
            for (int k = 0; k <= 6; k++) {
                if (s1[i][j] - k < 97)
                    s4[i][k] += (s1[i][j] - k + 26);
                else
                    s4[i][k] += (s1[i][j] - k);
            }
        }
    }
    for (int i = 1; i <= 3; i++) {          //反转         s1[i]是s的反转       即s2[i]=s
        for (int j = 1; j <= 3; j++) {      //后k个       s1[j]是s的后k个   即s4[j][k]=s
            for (int r = 1; r <= 3; r++) {  //前k个     s1[r]是s的前k个 即s3[r][k]=s
                for (int k = 0; k <= 6; k++) {
                    if (i != j && j != r && i != r && s2[i] == s4[j][k] && s2[i] == s3[r][k]) {
                        cout << s2[i];
                        return 0;
                    }
                }
            }
        }
    }
    return 0;
}

T2 考试时没想到从后往前遍历 直接暴力拿下60分 （为什么别人暴力都是清一色的30分…） 60分代码

#include <bits/stdc++.h>
using namespace std;
int n, m, q;
long long ans = 0;
int main() {
    freopen("board.in", "r", stdin);
    freopen("board.out", "w", stdout);
    cin >> n >> m >> q;
    bool a[n + 2][m + 2] = { 0 };
    for (int i1 = 1; i1 <= q; i1++) {
        int x, y, z;
        cin >> x >> y >> z;
        if (x === 0) {
            for (int i = 1; i <= m; i++) a[y][i] = z;
        }
        if (x === 1) {
            for (int i = 1; i <= n; i++) a[i][y] = z;
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (a[i][j] == 1)
                ans++;
        }
    }
    cout << ans;
    return 0;
}

AC做法 如果是染行就+m-列的个数，染列就+n-行的个数

#include <bits/stdc++.h>
using namespace std;
long long n,m,q,x[1000001],y[1000001],z[1000001],sum1=0,sum2=0,line1[1000001],line2[1000001];
long long ans=0;
int main() {
    freopen("board.in", "r", stdin);
    freopen("board.out", "w", stdout);
    cin>>n>>m>>q;
    for(int i=1;i<=q;i++)cin>>x[i]>>y[i]>>z[i];
    for(int i=q;i>=1;i--){
        if(x[i]==0){
            if(line1[y[i]]==1){
                continue;
            }
            else{
                if(z[i]==1){
                    ans+=m;
                    ans-=sum1; 
                } 
                line1[y[i]]=1;
                sum2++; 
            }
        }
        else{
            if(line2[y[i]]==1){
                continue;
            }
            else{
                if(z[i]==1){
                    ans+=n;
                    ans-=sum2;
                }
                line2[y[i]]=1;
                sum1++;
            }
        }
    }
    cout<<ans;
    return 0;
}

T3 数学题简单想一下如果遇到b就++，遇到a后有b就*2

#include <bits/stdc++.h>
using namespace std;
long long ans = 0, mod = 1e9 + 7, b1 = 0, t = 1;
int n, l[1000002], ll = 1;
int main() {
    freopen("string.in", "r", stdin);
    freopen("string.out", "w", stdout);
    string s;
    cin >> s;
    n = s.size();
    for (int i = n - 1; i >= 0; i--) {
        if (s[i] == &#039;b&#039;) {
            b1++;
        }
        if (s[i] == &#039;a&#039; && s[i + 1] == &#039;b&#039;) {
            l[ll] = b1 + l[ll - 1];
            s[i] = &#039;b&#039;;
            b1 = 0;
            ll++;
        }
    }
    for (int i = ll - 1; i >= 1; i--) {
        t %= mod;
        ans += ((t * l[i]) % mod);
        ans %= mod;
        t *= 2;
    }
    cout << ans % mod;
    return 0;
}

T4 改不出来