#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 }

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 }