#291 div.2

A.水题 数字翻转，将每一位大于等于5的数字t翻转成9-t，注意不要有前导0，且翻转后数字的位数不变（即9999->9000...刚开始以为应该翻转成0了= =）

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 using namespace std;

 6 int main()

 7 {

 8     char x[20];

 9     cin >> x;

10     int len = strlen(x);

11     for(int i = 0; i < len; i++)

12         if(x[i] >= '5') x[i] = '9' - x[i] + '0';

13     if(x[0] == '0') x[0] = '9';

14     cout << x << endl;

15     return 0;

16 }

17 #include<iostream>

18 #include<cstdio>

19 #include<cstdlib>

20 #include<cstring>

21 using namespace std;

22 int main()

23 {

24     char x[20];

25     cin >> x;

26     int len = strlen(x);

27     for(int i = 0; i < len; i++)

28         if(x[i] >= '5') x[i] = '9' - x[i] + '0';

29     if(x[0] == '0') x[0] = '9';

30     cout << x << endl;

31     return 0;

32 }

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B.水题 斜率，每输入一个坐标(xi,yi)求出该点与标准点连线的斜率并记入数组k[]中，排序（sort），去重（unique）；

         NOTE：斜率注意的基本问题---斜率为无穷大与0时的特判；

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 #include<algorithm>

 6 using namespace std;

 7 const int maxn = 1010;

 8 int main()

 9 {

10     int n, cnt1 = 0, cnt2 = 0;

11     double x0, y0, x, y;

12     double k[maxn];

13     scanf("%d%lf%lf", &n, &x0, &y0);

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

15     {

16         scanf("%lf%lf", &x, &y);

17         if(x-x0 == 0) {cnt1++; k[i] = 0; continue;}

18         if(y-y0 == 0) {cnt2++; k[i] = 0; continue;}

19         k[i] = (y-y0)/(x-x0);

20     }

21     sort(k, k+n);

22     int ans = unique(k, k+n) - k;

23     if(cnt1 && cnt2) ans++;

24     cout << ans << endl;

25     return 0;

26 }

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