ZZUACM 2015 暑假集训 round 01

A. Encoding

Problem Description

Given a string containing only ‘A’ - ‘Z’, we could encode it using the following method:

1. Each sub-string containing k same characters should be encoded to “kX” where “X” is the only character in this sub-string.

2. If the length of the sub-string is 1, ‘1’ should be ignored.

Input

The first line contains an integer N (1 <= N <= 100) which indicates the number of test cases. The next N lines contain N strings. Each string consists of only ‘A’ - ‘Z’ and the length is less than 10000.

Output

For each test case, output the encoded string in a line.

Sample Input

2
ABC
ABBCCC

Sample Output

ABC
A2B3C

题意：给你串字符串。让你统计并输出连续的字母，假设字母个数为1。则1忽略掉

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

int main()
{
    int n;
    scanf("%d", &n);
    getchar();
    while(n--){
        char str[10010];
        scanf("%s", str);
        int len = strlen(str);
        int count = 1;
        for(int i = 0; i < len; i++){
            if(str[i] == str[i+1])
                count++;
            else{
                if(count == 1)
                    printf("%c", str[i]);
                else{
                    printf("%d%c", count, str[i]);
                    count = 1;
                }
            }
        }
        printf("\n");
    }
    return 0;
}

B. Easier Done Than Said?

Problem Description

Password security is a tricky thing. Users prefer simple passwords that are easy to remember (like buddy), but such passwords are often insecure. Some sites use random computer-generated passwords (like xvtpzyo), but users have a hard time remembering them and sometimes leave them written on notes stuck to their computer. One potential solution is to generate “pronounceable” passwords that are relatively secure but still easy to remember.

FnordCom is developing such a password generator. You work in the quality control department, and it’s your job to test the generator and make sure that the passwords are acceptable. To be acceptable, a password must satisfy these three rules:

It must contain at least one vowel.

It cannot contain three consecutive vowels or three consecutive consonants.

It cannot contain two consecutive occurrences of the same letter, except for ‘ee’ or ‘oo’.

(For the purposes of this problem, the vowels are ‘a’, ‘e’, ‘i’, ‘o’, and ‘u’; all other letters are consonants.) Note that these rules are not perfect; there are many common/pronounceable words that are not acceptable.

Input

The input consists of one or more potential passwords, one per line, followed by a line containing only the word ‘end’ that signals the end of the file. Each password is at least one and at most twenty letters long and consists only of lowercase letters.

Output

For each password, output whether or not it is acceptable, using the precise format shown in the example.

Sample Input

a
tv
ptoui
bontres
zoggax
wiinq
eep
houctuh
end

Sample Output

is acceptable.
is not acceptable.
is not acceptable.
is not acceptable.
is not acceptable.
is not acceptable.
is acceptable.
is acceptable.

题意：题目中规定了password的形式例如以下：

1.password中必须有一个是元音字母

2.password中不能出现三个连续的元音字母或者三个连续的辅音字母

3.password中不能出现两个连续的同样的字母，除了”ee”和”oo”

给你一串password。让你推断这个password是否符合，输出按例子格式输出

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

int main()
{
    char code[25];
    while(scanf("%s", code)!=EOF){
        if(!strcmp(code, "end"))    
            break;
        int v, vv, cc, o, flag = 1;    //v标记是否有元音字母，vv是连续元音字母的个数，cc连续辅音字母的个数。o是否有两个反复的字母
        v = vv = cc = o = 0;
        int len = strlen(code);
        if(len == 1){       //仅仅有1个字母的情况单独推断
            if(code[0]==‘a‘ || code[0]==‘e‘ || code[0]==‘i‘ || code[0]==‘o‘ || code[0]==‘u‘)
                flag = 1;
        }
        else{
            for(int i = 0; i < len; i++){
                if(code[i]==‘a‘ || code[i]==‘e‘ || code[i]==‘i‘ || code[i]==‘o‘ || code[i]==‘u‘){
                    v = 1;
                    vv++;
                    cc = 0;
                }
                else{
                    vv = 0;
                    cc++;
                }
                if(code[i] == code[i+1] && code[i]!=‘e‘ && code[i]!=‘o‘)
                    o = 1;
                if(cc==3 || vv==3 || o==1){    //推断是否违反那几条规则
                    flag = 0;
                    break;
                }
            }
            if(!v)  flag = 0;     //推断是否含有元音字母
        }
        if(flag)
            printf("<%s> is acceptable.\n", code);
        else
            printf("<%s> is not acceptable.\n", code);
    }
    return 0;
}

D. A + B Problem II

Problem Description

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.

Output

For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line is the an equation “A + B = Sum”, Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.

Sample Input

2
1 2
112233445566778899 998877665544332211

Sample Output

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

题意：大数相加，用数组处理

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

int main()
{
    char str1[1010], str2[1010];
    int a[1010], b[1010], c[1010];
    int count = 1, n;
    scanf("%d", &n);
    getchar();
    while(n--){
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        memset(c, 0, sizeof(c));

        char str1[1010], str2[1010];
        scanf("%s %s", str1, str2);
        int len1 = strlen(str1);
        int len2 = strlen(str2);

        for(int i = 0; i < len1; i++)
            a[i] = str1[len1-i-1]-‘0‘;
        for(int i = 0; i < len2; i++)   
            b[i] = str2[len2-i-1]-‘0‘;

        int m = len1 > len2? len1:len2;
        int up = 0;
        for(int i = 0; i < m; i++){       //大数相加的核心
            c[i] = a[i]+b[i]+up;
            up = c[i]/10;
            c[i]%=10;
        }

        printf("Case %d:\n", count++);
        printf("%s + %s = ", str1, str2);
        int i = 0;
        for(i = 1001; c[i]==0; i--);
        for( ; i >= 0; i--)
            printf("%d", c[i]);
        printf("\n");
        if(n)   printf("\n");
    }
    return 0;
}

E. 素数判定

Problem Description

对于表达式n^2+n+41。当n在（x,y）范围内取整数值时（包括x,y）(-39<=x

Input

输入数据有多组，每组占一行，由两个整数x。y组成。当x=0,y=0时，表示输入结束。该行不做处理。

Output

对于每一个给定范围内的取值。假设表达式的值都为素数，则输出”OK”,否则请输出“Sorry”,每组输出占一行。

Sample Input

0 1
0 0

Sample Output

OK

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
int mark[2800];

void prime(){                      //先用筛选法把数素都标记好
    memset(mark, 1, sizeof(mark));
    mark[0] = mark[1] = 0;
    int n = sqrt(2800);
    for(int i = 2; i <= n; i++){
        if(mark[i]){
            for(int j = i*i; j < 2800; j+=i)
                mark[j] = 0;
        }
    }
}

int main()
{
    int x, y;
    prime();
    while(scanf("%d %d", &x, &y)!=EOF){
        if(!x && !y)
            break;
        int flag = 1;
        for(int i = x; i <= y; i++){
            if(mark[i*i+i+41])
                continue;
            else{
                flag = 0;
                break;
            }
        }
        if(flag)    printf("OK\n");
        else    printf("Sorry\n");
    }   
    return 0;
}

F. 蟠桃记

Problem Description

喜欢西游记的同学肯定都知道悟空偷吃蟠桃的故事，你们一定都认为这猴子太闹腾了，事实上你们是有所不知：悟空是在研究一个数学问题！
什么问题？他研究的问题是蟠桃一共同拥有多少个！

只是。到最后，他还是没能解决这个难题，呵呵^-^
当时的情况是这种：
第一天悟空吃掉桃子总数一半多一个，第二天又将剩下的桃子吃掉一半多一个，以后每天吃掉前一天剩下的一半多一个，到第n天准备吃的时候仅仅剩下一个桃子。

聪明的你，请帮悟空算一下，他第一天開始吃的时候桃子一共同拥有多少个呢？

Input

输入数据有多组，每组占一行，包括一个正整数n（1

Output

对于每组输入数据，输出第一天開始吃的时候桃子的总数。每一个測试实例占一行。

Sample Input

2
4

Sample Output

4
22

数学问题

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

int main()
{
    int n;
    while(scanf("%d", &n)!=EOF){
        int sum = 1;
        for(int i = 2; i <= n; i++ ){
            sum = sum*2+2;
        }
        printf("%d\n", sum);
    }
    return 0;
}

G. 大菲波数

Problem Description

Fibonacci数列，定义例如以下：
f(1)=f(2)=1
f(n)=f(n-1)+f(n-2) n>=3。

计算第n项Fibonacci数值。

Input

输入第一行为一个整数N。接下来N行为整数Pi（1<=Pi<=1000）。
Output
输出为N行。每行为相应的f(Pi)。

Sample Input

5
1
2
3
4
5

Sample Output

1
1
2
3
5

题意：还是大数相加，开个二维数组搞定

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int fib[1010][220];

void fibo(){
    for(int i = 0; i < 1010; i++)
        for(int j = 0; j < 220; j++)
            fib[i][j] = 0;

    fib[0][0] = fib[1][0] = 1;
    for(int i = 2; i < 1000; i++){
        int up, j;
        up = j = 0;
        while(j < 220){
            fib[i][j] = fib[i-1][j] + fib[i-2][j] + up;
            up = fib[i][j]/10;
            fib[i][j]%=10;
            j++;
        }
    }
}

int main()
{
    int n;
    fibo();
    scanf("%d", &n);
    while(n--){
        int m, i;
        scanf("%d", &m);

        for(i = 219; fib[m-1][i] == 0; i--);
        for(; i>=0; i--)
            printf("%d", fib[m-1][i]);
        printf("\n");
    }
    return 0;
}