【题解】 —算法竞赛入门经典第二版 【第三章例题】【6/6】

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例题 3-1

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#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;

int main()
{
    int flag=1;
    int c;
    while((c = getchar()) != EOF)
    {
        if(c == '"')
        {
            if(flag)
                cout << "``";
            else
                cout << "''";
            flag = !flag;
        }
        else
            printf("%c", c);
    }
    return 0;
}

例题 3-2

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#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

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;
}

例题 3-3

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#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

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

char r(char ch)
{
    if(ch >= 'A' && ch <= 'Z')
        return rev[ch - 'A'];
    else
        return rev[ch - '0' + 25];
}
int main()
{
    char s[30];
    while(~scanf("%s", s))
    {
        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;
            if(r(s[i]) != s[len-1-i])   m = 0;
        }
        printf("%s -- is %s.\n\n", s, res[m * 2 + p]);
    }
    return 0;
}

例题 3-4

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#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAX = 1010;

int n, cnt1, cnt2;
int a[MAX], b[MAX];
int main()
{
    for(int cas = 1; scanf("%d",&n) && n; ++cas)
    {
        printf("Game %d:\n", cas);
        for(int i = 0; i < n; ++i)
            scanf("%d",&a[i]);
        while(true)
        {
            cnt1 = cnt2 = 0;
            for(int i = 0; i < n; ++i)
            {
                scanf("%d",&b[i]);
                if(a[i] == b[i])
                    ++cnt1;
            }
            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;
                }
                cnt2 += min(c1, c2);
            }
            printf("    (%d,%d)\n", cnt1, cnt2 - cnt1);
        }
    }
    return 0;
}

例题 3-5

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#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAX = 100010;

int t, n;
int res[MAX];
void init()
{
    for(int i = 1; i < MAX; ++i)
    {
        int x = i, y = i;
        while(x)
        {
            y += x % 10;
            x /= 10;
        }
        if(res[y] == 0 || i < res[y])   res[y] = i;
    }
}
int main()
{
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("%d\n", res[n]);
    }
    return 0;
}

例题 3-6

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#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

int les(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()
{
    char s[105];
    int n, ans, len;
    while(~scanf("%d",&n))
    {
        while(n--)
        {
            ans = 0;
            scanf("%s",s);
            len = strlen(s);
            for(int i = 0; i < len; ++i)
                if(les(s, i, ans))
                    ans = i;
            for(int i = 0; i < len; ++i)
                cout << s[(ans+i)%len];
            cout << endl;
        }
    }
    return 0;
}
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