Box of Bricks

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 9216 Accepted Submission(s): 2157

Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one upon another and builds stacks of different height. “Look, I\'ve built a wall!”, he tells his older sister Alice. “Nah, you should make all stacks the same height. Then you would have a real wall.”, she retorts. After a little consideration, Bob sees that she is right. So he sets out to rearrange the bricks, one by one, such that all stacks are the same height afterwards. But since Bob is lazy he wants to do this with the minimum number of bricks moved. Can you help?

Input The input consists of several data sets. Each set begins with a line containing the number n of stacks Bob has built. The next line contains n numbers, the heights hi of the n stacks. You may assume 1≤n≤50 and 1≤hi≤100. The total number of bricks will be divisible by the number of stacks. Thus, it is always possible to rearrange the bricks such that all stacks have the same height. The input is terminated by a set starting with n = 0. This set should not be processed.

Output             For each set, print the minimum number of bricks that have to be moved in order to make all the stacks the same height. Output a blank line between each set.

Sample Input 6 5 2 4 1 7 5 0

Sample Output Set #1 The minimum number of moves is 5.

求平均數，計算大於或者小於平均數的所有值與平均數的差值的和。

#include <bits/stdc++.h>
using namespace std;
int a[100];

int main()
{
    int n, aver, k = 0;
    while(cin >> n && n != 0)
    {
        k++;
        int sum = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            sum += a[i];
        }
        aver = sum / n;
        sum = 0;
        for(int i = 0; i < n; i++)
        {
            if(a[i] > aver)
            {
                sum += (a[i] - aver);
            }
        }
        printf("Set #%d\n", k);
        printf("The minimum number of moves is %d.\n\n", sum);
    }

    return 0;
}

3
2005 25
1855 12
2004 10000

2108
1904
43236

計算出從這年開始（包括這一年）的第一個閏年，然後開始向後以加四的形式查詢，剔除非閏年的存在，達到一定個數即輸出年份。

#include <bits/stdc++.h>
using namespace std;
int a[100];

bool isleap(int y)
{
    if(y % 400 == 0 || (y % 4 == 0 && y % 100 != 0))
    {
        return true;
    }
    else
    {
        return false;
    }
}

int main()
{
    int n;
    cin >> n;
    while(n--)
    {
        int y, b;
        while(cin >> y >> b)
        {
            while(y)
            {
                if(isleap(y))
                {
                    b--;
                    break;
                }
                else
                {
                    y++;
                }
            }
            int k = 0;
            while(b)
            {
                y += 4;
                if(isleap(y))
                {
                    b--;
                }
            }
            cout << y << endl;
        }
    }
    return 0;
}

5
1 1 3 2 2
3
1 2 1
0

3
2

題目的意思我不知道是指要找出來的是那個數是奇數個還是隻有一個，但是解法是一樣的。使用異或的性質。不斷異或，最後得到的就是出現奇數次的那個數。

也可以使用set，出現一次插入，再次出現刪去，最後僅剩的一個就是所求的。

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int n;
    int a, b;
    while(cin >> n && n != 0)
    {
        scanf("%d", &a);
        for(int i = 1; i < n; i++)
        {
            scanf("%d", &b);
            a = a ^ b;
        }
        cout << a << endl;
    }
    return 0;
}

#include <set>
#include <stdio.h>
using namespace std;
int main()
{
    int n,x;
    set <int> S;
    while(scanf("%d",&n),n)
    {
        while(n--)
        {
            scanf("%d",&x);
            if(S.find(x) == S.end())    
                S.insert(x);
            else                        
                S.erase(x);
        }
        printf("%d\n",*S.begin());
        S.clear();
    }
    return 0;
}

2
3
4

7
6

開始的做法是先找出最後一位。每次乘完一次或者兩次或者三次，就去除最低一位。但是都超時了。

最後是用找規律的方法：找出末位，1.5.6.0都是怎麼乘都等於本身，而其他的都是四個一迴圈。於是便可以得出結果了。（是不是很像高中的程式框圖）

#include <bits/stdc++.h>
using namespace std;
int getnum(int n)
{
    while(n >= 10)
    {
        n = n % 10;
    }
    return n;
}

int pow1(int a, int b)
{
    if(b == 0)
    {
        b = 4;
    }
    int res = 1;
    for(int i = 0; i < b; i++)
    {
        res = res * a;
    }
    res = getnum(res);
    return res;
}

int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        int n;
        cin >> n;
        int m = getnum(n);
        int res1;
        if(m == 0 || m == 1 || m == 5 || m == 6)
        {
            cout << m << endl;
        }
        else
        {
            m = pow1(m, n%4);
            cout << m << endl;
        }
    }
    return 0;
}

Buildings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2791 Accepted Submission(s): 1979

Problem Description We divide the HZNU Campus into N*M grids. As you can see from the picture below, the green grids represent the buidings. Given the size of the HZNU Campus, and the color of each grid, you should count how many green grids in the N*M grids.

Input Standard input will contain multiple test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. The first line of each test case contains two integers n and m(1<=n,m<=100), the size of the campus. Then follow n lines, each line containing m integers. The j-th integer in the i-th line is the color of that grid, 0 stands for white color, while 1 stands for green.

Output             Results should be directed to standard output. For each case, output an integers T, the total green grids in the N*M size campus.

Sample Input 2 2 2 1 1 0 0 3 3 1 0 1 0 0 1 1 1 0

Sample Output 2 5

列舉，找到一的個數即可。

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int m, n;
        scanf("%d%d", &m, &n);
        int sum = 0, x;
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < n; j++)
            {
                scanf("%d", &x);
                if(x == 1)
                {
                    sum++;
                }
            }
        }
        printf("%d\n", sum);
    }

    return 0;
}

4
+ 1 2
- 1 2
* 1 2
/ 1 2

3
-1
2
0.50

開始一直出錯，是直接將除法當作是沒有保留小數點的計算，忽略了可能出現取整的情況。同時printf的保留是四捨五入的。

#include <bits/stdc++.h>
using namespace std;

int main()
{
    char oper;
    int a, b;
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%s", &oper);
        if(oper != '/')
        {
            scanf("%d%d", &a, &b);
            if(oper == '+')
            {
                printf("%d\n", a+b);
            }
            if(oper == '-')
            {
                printf("%d\n", a-b);
            }
            if(oper == '*')
            {
                printf("%d\n", a*b);
            }
        }
        else
        {
            int a, b;
            scanf("%d%d", &a, &b);
            if(a/b*b == a)
            {
                printf("%d\n", a/b);
            }
            else
            {
                printf("%.2f\n", double(a)/double(b));
            }
        }
    }
    return 0;
}

A B C D F
B F F C C A
D C E F

2.00
1.83
Unknown letter grade in input

卡的點在於%.2f寫成了%.2F。沒有難度，直接掃。

#include <bits/stdc++.h>
using namespace std;

int main()
{
    string str;
    while(getline(cin, str))
    {
        double sum = 0;
        int flag = 0;
        int n = str.size();
        double k = (n+1)/2;
        for(int i = 0; i < n; i = i + 2)
        {
            if(str[i] == 'A')
            {
                sum += 4;
            }
            else if(str[i] == 'B')
            {
                sum += 3;
            }
            else if(str[i] == 'C')
            {
                sum += 2;
            }
            else if(str[i] == 'D')
            {
                sum += 1;
            }
            else if(str[i] == 'F')
            {
                sum += 0;
            }
            else
            {
                flag = 1;
            }
        }
        if(flag == 0)
        {
            printf("%.2f\n", sum/k);
        }
        else
        {
            printf("Unknown letter grade in input\n");
        }
    }
    return 0;
}

2
HAL
SWERC

String #1
IBM

String #2
TXFSD

直接利用轉化關係，但是要注意的是轉化時候 Z是轉化成A的，而且轉化那裡只能寫成相互形式，而不能是+‘1’

<pre name="code" class="cpp">#include <bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    scanf("%d", &t);
    char s[100];
    for(int i = 1; i <= t; i++)
    {
        cin >> s;
        for(int j = 0; j < strlen(s); j++)
        {
            if(s[j] == 'Z')
            {
                s[j] = 'A';
            }
            else
            {
                s[j] = s[j] + ('I' - 'H');
            }
        }
        printf("String #%d\n", i);
        printf("%s\n\n", s);
    }
    return 0;
}