数学 2016.2.16

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#数学

------------数学------------- 同时被 2 个专栏收录

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本文解析了五道UVa在线编程题目的解题思路与实现代码，包括棋盘上的蚂蚁、立方体着色、特定数学问题求解、蜗牛爬井问题及计算步数等，展示了使用C++解决实际问题的方法。

1、UVa 10161 Ant on a Chessboard(棋盘上的蚂蚁)

#include <iostream>
#include <cmath>

using namespace std;

int IsSquares(int n);

int main()
{
    int N;

    while (cin>>N && N!=0) {
        int i = N;
        while (!IsSquares(i)) {
            ++i;
        }
        int x, y;
        int Sqrti = sqrt(i);
        if (Sqrti%2 == 0) {
            if (N-(Sqrti-1)*(Sqrti-1) <= Sqrti) {
                x = N - (Sqrti-1)*(Sqrti-1);
                y = Sqrti;
            } else {
                x = Sqrti;
                y = i + 1 - N;
            }
        } else {
            if (N-(Sqrti-1)*(Sqrti-1) <= Sqrti) {
                x = Sqrti;
                y = N-(Sqrti-1)*(Sqrti-1);
            } else {
                x = i + 1 - N;
                y = Sqrti;
            }
        }
        cout<<x<<" "<<y<<endl;
    }
    return 0;
}

int IsSquares(int n)
{
    if (floor(sqrt(n)) == sqrt(n)) {
        return 1;
    } else {
        return 0;
    }
}

2、UVa 253 Cube painting(立方体着色)

#include <iostream>

using namespace std;

int main()
{
    char s[20];
    char color1[10], color2[10];

    while (cin>>s) {
        int i, j;
        int a1, a2, a3, b1, b2, b3;
        for (j=0,i=1; i<=6; ++i,++j) {
            color1[i] = s[j];
//            cout<<color1[i];
        }
//        cout<<endl;
        a1 = color1[1] + color1[6];
        a2 = color1[2] + color1[5];
        a3 = color1[3] + color1[4];
//        cout<<a1<<" "<<a2<<" "<<a3<<endl;
        for (; i<=12; ++i,++j) {
            color2[i-6] = s[j];
//            cout<<color2[i-6];
        }
//        cout<<endl;
        b1 = color2[1] + color2[6];
        b2 = color2[2] + color2[5];
        b3 = color2[3] + color2[4];
//        cout<<b1<<" "<<b2<<" "<<b3<<endl;
        int flag = 0;
        if (a1==b1 && a2==b2 && a3==b3) {
            flag = 1;
        } else if (a1==b1 && a2==b3 && a3==b2) {
            flag = 1;
        } else if (a1==b2 && a2==b1 && a3==b3) {
            flag = 1;
        } else if (a1==b2 && a2==b3 && a3==b1) {
            flag = 1;
        } else if (a1==b3 && a2==b1 && a3==b2) {
            flag = 1;
        } else if (a1==b3 && a2==b2 && a3==b1) {
            flag = 1;
        }
        if (flag) {
            cout<<"TRUE"<<endl;
        } else {
            cout<<"FALSE"<<endl;
        }
    }
    return 0;
}

3、UVa 10025 The ? 1 ? 2 ? … ? n = k problem

#include <iostream>

using namespace std;

int main()
{
    int T;

    while (cin>>T) {
        int k;
        while (T--) {
            cin>>k;
            if (k < 0) {
                k = -k;
            }
            int i = 1;
            int count = 0;
            while (1) {
                ++count;
                if (k-i == 0) {
                    break;
                }
                if (k-i < 0) {
                    if ((i-k)%2 == 0) {
                        break;
                    } else {
                        if (i%2 == 0) {
                            count += 1;
                            break;
                        } else {
                            count += 2;
                            break;
                        }
                    }
                }
                k -= i;
                ++i;
            }
            cout<<count<<endl;
            if (T > 0) {
                cout<<endl;
            }
        }
    }
    return 0;
}

4、UVa 573 The Snail(蜗牛)
题意：井的高度为H,蜗牛白天可以爬的高度为U,晚上滑落的高度为D,疲劳系数为F%,开始的时候蜗牛在井底，输出它在第几天会成功或失败
解题思路：
(1)、蜗牛第n个白天爬的高度为U-F%(n-1)(注意U-F%(n-1)<=0时，蜗牛白天会在原地不动)，每天下滑高度为D
(2)、某一天上爬后的高度>H则成功，下滑后的高度<0则失败

#include <iostream>

using namespace std;

int main()
{
    int H, U, D, F;


    while (cin>>H>>U>>D>>F && H!=0) {
        int day = 1;
        double Init_H = 0, Dis_Climb = U, H_After_Climb = U, H_After_Slid = U-D;
//        cout<<Init_H<<" "<<Dis_Climb<<" "<<H_After_Climb<<" "<<H_After_Slid<<endl;
        while ((H_After_Climb<=H) && (H_After_Slid>=0)) {
            Init_H = H_After_Slid;
            if (Dis_Climb-U*F/100.0 > 0) {
                Dis_Climb -= (U*F/100.0);
            } else {
                Dis_Climb = 0;
            }
            H_After_Climb = Init_H + Dis_Climb;
            H_After_Slid = H_After_Climb - D;
            ++day;
//            cout<<Init_H<<" "<<Dis_Climb<<" "<<H_After_Climb<<" "<<H_After_Slid<<endl;
        }
        if (H_After_Climb > H) {
            cout<<"success on day "<<day<<endl;
        } else if (H_After_Slid < 0) {
            cout<<"failure on day "<<day<<endl;
        }
    }
    return 0;
}

5、UVa 846 Steps(步数)

#include <iostream>

using namespace std;

int main()
{
    int n;

    while (cin>>n) {
        unsigned long long x, y;
        while (n--) {
            cin>>x>>y;
            int sum = y - x;
            int i = 2;
            int steps = 0;
            while (sum) {
                if (sum-i < 0) {
                    if (sum-(i/2) > 0) {
                        steps += 2;
                        break;
                    } else {
                        steps += 1;
                        break;
                    }
                }
                sum -= i;
                steps += 2;
                i += 2;
            }
            cout<<steps<<endl;
        }
    }
    return 0;
}