TEX Quotes

#include <stdio.h>

int main() {
    int c, q = 1;

    while ((c = getchar()) != EOF) {
        if (c == '"') {
            printf("%s", q ? "``" : "''");
            q = !q;
        } else
            printf("%c", c);
    }

    return 0;
}

WERTYU

#include <stdio.h>

char s[] = "`1234567890-=QWERTYUIOP[]\\ASDFGHJKL;'ZXCVBNM,./";

int main() {
    int i, c;

    while ((c = getchar()) != EOF) {
        for (i = 1; s[i] && s[i] != c; i ++ );

        if (s[i])
            putchar(s[i - 1]);
        else
            putchar(c);
    }

    return 0;
}

Palindromes

#include <stdio.h>
#include <string.h>
#include <ctype.h>

const char *rev = "A   3  HIL JM O   2TUVWXY51SE Z  8 ";
const char *msg[] = {"not a palindrome", "a regular palindrome", "a mirrored string", "a mirrored palindrome"};

char r(char ch) {
    if (isalpha(ch))
        return rev[ch - 'A'];
    return rev[ch - '0' + 25];
}

int main() {
    char s[30];

    while (scanf("%s", s) == 1) {
        int len = strlen(s);
        int p = 1, m = 1;

        for (int i = 0; i < (len + 1) / 2; i ++ ) {

            if (s[i] != s[len - 1 - i])
                p = 0; // 不是回文串

            if (r(s[i]) != s[len - 1 - i])
                m = 0; // 不是镜像串
        }

        printf("%s -- is %s.\n\n", s, msg[m * 2 + p]);
    }

    return 0;
}

Master-Mind Hints

#include <stdio.h>

#define maxn 1010

int main() {
    int n, a[maxn], b[maxn];
    int kase = 0;

    while (scanf("%d", &n) == 1 && n) {
        printf("Game %d:\n", ++ kase);

        for (int i = 0; i < n; i ++ )

            scanf("%d", &a[i]);

        for (;;) {

            int A = 0, B = 0;

            for (int i = 0; i < n; i ++ ) {

                scanf("%d", &b[i]);

                if (a[i] == b[i])
                    A ++ ;
            }

            if (b[0] == 0)
                break;

            for (int d = 1; d <= 9; d ++ ) {

                int c1 = 0, c2 = 0;

                for (int i = 0; i < n; i ++ ) {

                    if (a[i] == d)
                        c1 ++ ;

                    if (b[i] == d)
                        c2 ++ ;
                }

                if (c1 < c2)
                    B += c1;
                else
                    B += c2;
            }

            printf("    (%d,%d)\n", A, B - A);
        }
    }

    return 0;
}

Digit Generator

#include <stdio.h>
#include <string.h>

#define maxn 100005

int ans[maxn];

int main() {
    int T, n;
    memset(ans, 0, sizeof ans);

    for (int m = 1; m < maxn; m ++ ) {

        int x = m, y = m;

        while (x > 0) {
            y += x % 10;
            x /= 10;
        }

        if (ans[y] == 0 || m < ans[y])
            ans[y] = m;
    }

    scanf("%d", &T);

    while (T -- ) {
        scanf("%d", &n);
        printf("%d\n", ans[n]);
    }

    return 0;
}

Circular Sequence

#include <stdio.h>
#include <string.h>

#define maxn 105

int less(const char *s, int p, int q) {
    int n = strlen(s);

    for (int i = 0; i < n; i ++ )

        if (s[(p + i) % n] != s[(q + i) % n])
            return s[(p + i) % n] < s[(q + i) % n];
    return 0;
}

int main() {
    int T;
    char s[maxn];
    scanf("%d", &T);

    while (T -- ) {
        scanf("%s", s);
        int ans = 0;
        int n = strlen(s);

        for (int i = 1; i < n; i ++ )

            if (less(s, i, ans))
                ans = i;

        for (int i = 0; i < n; i ++ )

            putchar(s[(i + ans) % n]);
        putchar('\n');
    }

    return 0;
}

Score

#include <algorithm>
#include <iostream>

using namespace std;

char str[90];
int  score[90];

int main() {
    int n;
    scanf("%d", &n);

    while (n --) {
        scanf("%s", str);
        score[0] = (str[0] == 'O');

        for (int i = 1 ; str[i] ; ++ i) {

            score[i] = score[i - 1] + 1;

            if (str[i] == 'X')
                score[i] = 0;
        }

        int sum = 0;

        for (int i = 0 ; str[i] ; ++ i)

            sum += score[i];

        printf("%d\n", sum);
    }

    return 0;
}

Molar mass

#include <stdio.h>
#include <string.h>
#include <ctype.h> 

double mass (char c){
    switch(c){
        case 'C':
            return 12.01;
        case 'H':
            return 1.008;
        case 'O':
            return 16.00;
        case 'N':
            return 14.01;
    }
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        char s[85];
        double molar = 0.0;
        scanf("%s", s);
        double x = 0.0;
        int len = strlen(s);
        for(int i = 0; i < len; i++){
            if(isalpha(s[i])){
                x = mass(s[i]);
                molar += x;
            }
            else if(isdigit(s[i])){
                int n = s[i] - '0';
                if(isdigit(s[i+1])){
                    n = n*10 + (s[i+1]-'0');
                    i++;
                }  
                molar += (n-1)*x;
                x = 0.0;
            }       
        }
        printf("%.3f\n", molar);
    }
    return 0;
} 

Digit Counting

#include <bits/stdc++.h>        //万能头文件
using namespace std;

int main() {
    int n;
    cin >> n;

    while (n--) {
        stringstream ss ;       //实现整型、字符型的相互转化
        int num;
        cin >> num;
        char a[40000] ;

        for (int i = 1; i <= num; i++)

            ss << i;
        ss >> a ;
        int len = strlen(a) ;                   //字符数组长度，

        for (char i = '0'; i <= '9'; i++)

            cout << count(a, a + len, i) << (i == '9' ? "\n" : " ");    //字符0-9在数组中出现的次数
    }

    return 0 ;
}

Periodic Strings

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
    int t; cin >> t;
    char s[85];
    while (t--)
    {
        memset(s, 0, sizeof(s));
        scanf("%s", s);
        int len = strlen(s);
        int j;
        for (int i = 1; i <= len; ++i)
            if (len%i == 0) 
            {
                for (j = i; j <= len; j++)
                    if (s[j] != s[j%i])
                        break;
                if (j == len)
                {
                    cout << i << endl;
                    break;
                }
            }
        if (t)cout << endl;
    }
}

Puzzle

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

char puzzle[10][10];

void printPuzzle() {
    for (int i = 0; i < 5; ++i) {

        for (int j = 0; j < 4; ++j)

            cout << puzzle[i][j] << ' ';
        cout << puzzle[i][4] << endl;
    }
}

int main() {
    int kase = 0;

    while (fgets(puzzle[0], 7, stdin)) { //读入7-1=6个字符，即每行的5个字符加换行
        if (puzzle[0][0] == 'Z')
            break;

        for (int i = 1; i < 5; ++i)

            fgets(puzzle[i], 7, stdin);
        //cout << "Read Finished" << endl;
        //printPuzzle(); //测试读入情况
        int ei = 0, ej = 0; //找出空的那个格子

        for (int i = 0; i < 5; ++i)

            for (int j = 0; j < 5; ++j)

                if (puzzle[i][j] == ' ') {
                    ei = i, ej = j;
                    break;
                }

        bool valid = true; //标志非法指令出现过
        char moves[101];
        int cnt = 0;
        char c;

        while (cin >> c && c != '0') //考虑隔行读入
            moves[cnt++] = c;
        int ni = ei, nj = ej;

        for (int i = 0; i < cnt; ++i) {

            switch (moves[i]) {
                case 'A':
                    ni = ei - 1;
                    nj = ej;
                    break;

                case 'B':
                    ni = ei + 1;
                    nj = ej;
                    break;

                case 'L':
                    nj = ej - 1;
                    ni = ei;
                    break;

                case 'R':
                    nj = ej + 1;
                    ni = ei;
                    break;

                default:
                    break;
            }

            //非法情况:
            if (ni < 0 || ni > 4 || nj < 0 || nj > 4) {
                valid = false;
                break;
            } else {
                swap(puzzle[ei][ej], puzzle[ni][nj]);
                ei = ni, ej = nj; //更新空格位置
            }
        }

        getchar(); //非常重要！吞掉指令结束符0后面的回车！

        if (kase++)
            cout << endl; //非首个测试前输出换行
        cout << "Puzzle #" << kase << ":\n";

        if (valid == false)
            cout << "This puzzle has no final configuration." << endl;
        else
            printPuzzle();
    }

    return 0;
}

Crossword Answers

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int m,n,u=0;
    while((~scanf("%d",&m))&&m)
    {
        scanf("%d",&n);getchar();
        if(u) printf("\n");
        char puzzle[m][n+1];
        int i,j;
        for(i=0;i<m;i++)
            for(j=0;j<=n;j++)
                puzzle[i][j]=getchar();
        int num[m][n],k=0;
        for(i=0;i<m;i++)
            for(j=0;j<n;j++)
            {
                if(puzzle[i][j]!='*'&&(!i||!j||puzzle[i-1][j]=='*'||puzzle[i][j-1]=='*'))
                    num[i][j]=++k;
                else if(puzzle[i][j]!='*')
                    num[i][j]=-1;
                else
                    num[i][j]=0;
            }
//        for(i=0;i<m;i++)
//            for(j=0;j<n;j++)
//         {
//             printf("%d",num[i][j]);
//             printf("\n");
//         }
        printf("puzzle #%d:\n",++u);
        printf("Across\n");
        for(i=0;i<m;i++)
            for(j=0;j<n;j++)
            {
                if(num[i][j]>0)
                {
                    printf("%3d.",num[i][j]);
                    for(;j<n&&num[i][j];printf("%c",puzzle[i][j++]));
                    printf("\n");
                }

            }
        printf("Down\n");
        int ii;
        for(i=0;i<m;i++)
            for(j=0;j<n;j++)
            {
                if(num[i][j]>0)
                {
                    printf("%3d.",num[i][j]);
                    for(ii=i;ii<m&&num[ii][j];printf("%c",puzzle[ii][j]),num[ii++][j]=0);
                    printf("\n");
                }
            }
    }
    return 0;
}

DNA Consensus String

#include<bits/stdc++.h>

using namespace std;



string temp;

vector<string> v;         ///存储各个字符串

int n,len,T;

int cnt[26];                   ///记录26个英文字母各个所出现的次数



int main(){

   cin >> T;

   while(T--){

       v.clear();

       memset(cnt,0,sizeof(cnt));

       int sum = 0;

       cin >> n >> len;

       for(int i = 0;i < n;++i){         ///输入

           cin >> temp;

           v.push_back(temp);

       }

       for(int i = 0;i < len;++i){             ///记录每个字母出现的次数

           memset(cnt,0,sizeof(cnt));

           for(int j = 0;j < n;++j){

                char ch = v[j][i];

                cnt[ch-'A']++;

           }

           int ans = (max_element(cnt,cnt+26)-cnt); ///求出出现次数最多字母的下标

           printf("%c",ans+'A');                    ///输出这个字符

           char ch = ans+'A';

           for(int j = 0;j < n;++j){                ///累计汉明距离

                if(v[j][i]!=ch)

                    sum++;

           }

       }

       printf("\n");

       printf("%d\n",sum);

    }

   return 0;

}

Repeating Decimals

//#define LOCAL
#include <stdio.h>
#include <string.h>
const int maxn = 100000 + 5;

int a, b;
int vis[maxn], res[maxn];

int main() {
#ifdef LOCAL
    freopen("data.in", "r", stdin);
    freopen("data.out", "w", stdout);
#endif
    while(scanf("%d%d", &a, &b) == 2) {
        memset(vis, -1, sizeof(vis));
        int c = a % b, cnt = 0;
        c *= 10;
        while(vis[c] == -1) {
            res[cnt] = c / b;
            vis[c] = cnt++;
            c %= b;
            c *= 10;
        }
        // repeating cycle start-position
        int sta_pos = vis[c];
        printf("%d/%d = %d.", a, b, a/b);
        for(int i = 0; i < sta_pos; i++) {
            printf("%d", res[i]);
        }
        printf("(");
        if(cnt - sta_pos <= 50) {
            for(int i = sta_pos; i < cnt; i++) {
                printf("%d", res[i]);
            }
        } else {
            for(int i = sta_pos; i < sta_pos+50; i++) {
                printf("%d", res[i]);
            }
            printf("...");
        }
        printf(")\n");
        printf("   %d = number of digits in repeating cycle\n\n", cnt - sta_pos);
    }
    return 0;
}

All in All

#include <stdio.h>
#include <string.h>

char a[100005], b[100005];
int main() {
    while (~scanf("%s%s", a, b)) {
        int star = 0, lenb = strlen(b), lena = strlen(a);
        for (int i = 0; i < lenb; i ++) {
            if (a[star] == b[i])
                star ++;
            if (star == lena) {
                printf("Yes\n");
                break;
            }
        }
        if (star != lena)
            printf("No\n");
    }
    return 0;
}

Box

#include<stdio.h>
#include<algorithm>
using namespace std;
pair <int,int> box[6];//注意书写方式
int i;
int main(){
    while(1){
        for(i=0;i<6;i++){
            if(scanf("%d%d",&box[i].first,&box[i].second)!=2) return 0;
            if(box[i].first>box[i].second)
                swap(box[i].first,box[i].second);
        }
        sort(box,box+6);//我试了一下，此处排序是按照先first递增，其次second递增排序
        puts(box[0].first==box[1].first&&box[1].first==box[2].first&&box[2].first==box[3].first&&
             box[0].second==box[1].second&&box[1].second==box[4].first&&box[4].first==box[5].first&&
             box[2].second==box[3].second&&box[3].second==box[4].second&&box[4].second==box[5].second?
             "POSSIBLE":"IMPOSSIBLE");
    }
}

Kickdown

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main(){
    char a[105],b[105];
    int i,j,l1,l2;
    while(scanf("%s%s",a,b)!=EOF){
        int lena=strlen(a);
        int lenb=strlen(b);
        for(i=0;i<lena;i++){ //以字符串a为基准，b滑动匹配 
            int flag=1;
            for(j=0;j<lenb && (i+j)<lena;j++){  //依次对比 
                if((a[i+j]=='2') && (b[j]=='2')){ //不匹配，b往后滑，再依次匹配 
                    flag=0;
                    break;
                }
            }
            if(flag){
                break;
            }
        }
        l1=max(lena,lenb+i); //取（基准字符串a的长度）与（滑动字符串b的长度加滑动长度）中最大值，保证能够涵盖匹配后的字符串 

        for(i=0;i<lenb;i++){
            int flag=1;
            for(j=0;j<lena && (i+j)<lenb;j++){
                if((b[i+j]=='2') && (a[j]=='2')){
                    flag=0;
                    break;
                }
            }
            if(flag){
                break;
            }
        }
        l2=max(lenb,lena+i); 

        int l=min(l1,l2); //保证求出的是最短容器长度 
        printf("%d\n",l);
    }
    return 0;
} 

Floating-Point Numbers

#include <iostream>
#include <sstream>
#include <string>
#include <cmath>

using namespace std;

int main() {
    double M[20][40];
    long long E[20][40];

    // 打表
    for(int i = 0; i <= 9; ++i) for(int j = 1; j <= 30; ++j) {
        double m = 1 - pow(2, -1 - i), e = pow(2, j) - 1;
        double t = log10(m) + e * log10(2);
        E[i][j] = t, M[i][j] = pow(10, t - E[i][j]);
    }

    // 输入并输出结果
    string in;
    while(cin >> in && in != "0e0") {
        // 处理输入
        for(string::iterator i = in.begin(); i != in.end(); ++i) if(*i == 'e') *i = ' ';
        istringstream ss(in);
        double A; int B;
        ss >> A >> B;
        while(A < 1) A *= 10, B -= 1;
        // 在打好的表中寻找答案
        for(int i = 0; i <= 9; ++i) for(int j = 1; j <= 30; ++j) {
            if(B == E[i][j] && (fabs(A - M[i][j]) < 1e-4 || fabs(A / 10 - M[i][j]) < 1e-4)) {
                cout << i << ' ' << j << endl;
                break;
            }
        }
    }
}