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A. Chewbaсca and Number
题意：
Inverting digit t means replacing it with digit 9 - t. Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn‘t contain leading zeroes. 分析： It is obvious that all the digits, which are greater than 4, need to be inverted. The only exception is 9, if it‘s the first digit. 时间复杂度：
O(len)
#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; char s[ 1000 ]; int main() { scanf( "%s", s ); int i = 0; if( s[i] == ‘9‘ ) i++; for( ; s[i]; ++i ) { char c = ‘9‘ - s[i] + ‘0‘; if( c < s[i] ) s[i] = c; } puts( s ); return 0; }
B. Han Solo and Lazer Gun
题意:
在一个坐标系上有一个机关枪，能够射穿一条线上的所有敌人，问射死整个坐标系中的敌人，至少需要多少发子弹。
分析：
开始考虑的是计算斜率，但是却用了int型导致WA两次，后来改为double AC了，但是使用这样的斜率是存在无常的。如果数据大一点也许就会出现错误。
这里可以考虑三点共线的条件：
Points (x1, y1), (x2, y2), (x3, y3) are on the same line ,if (x2 - x1)(y3 - y1) = (x3 - x1)(y2 - y1).
或者使用gcd和set来避免精度问题
#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; int sx, sy, n; double res[ 1010 ]; int main() { scanf( "%d %d %d", &n, &sx, &sy ); int flag = 0; int ct = 0; int out = 0; for( int i = 0; i < n; ++i ) { int x, y; scanf( "%d %d", &x, &y ); if( x == sx && y == sy ) continue; int ex = x - sx; int ey = y - sy; if( ex == 0 ) { int k = 0x7fffffff; res[ ct++ ] = k; } else if( ey == 0 ) { int k = 0x3f3f3f3f; res[ ct++ ] = k; } else { double k = (double)ex / (double)ey; res[ ct++ ] = k; } } sort( res, res + ct ); for( int i = 0; i < ct; ++i ) { if( res[i] == res[i+1] ) continue; out += 1; } printf( "%d\n", out ); return 0; }
#include <bits/stdc++.h> using namespace std; const int N = 1000; long long x[N]; long long y[N]; bool killed[N]; int main() { int n, x0, y0; cin >> n >> x0 >> y0; for (int i = 0; i < n; ++i) { cin >> x[i] >> y[i]; x[i] -= x0; y[i] -= y0; } int result = 0; for (int i = 0; i < n; ++i) if (!killed[i]) { for (int j = 0; j < n; ++j) if (x[i] * y[j] - y[i] * x[j] == 0) killed[j] = true; ++result; } cout << result << endl; return 0; }
#include <cstdio> #include <algorithm> #define N 1005 #define fi(a, b, c) for(int a = (b); a < (c); a++) #define fd(a, b, c) for(int a = (b); a > (c); a--) #define FI(a, b, c) for(int a = (b); a <= (c); a++) #define FD(a, b, c) for(int a = (b); a >= (c); a--) #define fe(a, b, c) for(int a = (b); a; a = c[a]) using namespace std; int n, x, y, a[N], b[N], ans; bool u[N]; int main(){ scanf("%d %d %d", &n, &x, &y); fi(i, 0, n) scanf("%d %d", &a[i], &b[i]); fi(i, 0, n) if(!u[i]){ ans++; fi(j, 0, n) if((a[i] - x) * (b[j] - y) == (a[j] - x) * (b[i] - y)) u[j] = 1; } printf("%d\n", ans); scanf("\n"); }
#include <iostream> #include <vector> #include <set> using namespace std; int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a % b); } int main() { ios::sync_with_stdio(false); int N, X0, Y0; cin >> N >> X0 >> Y0; set< pair<int, int> > S; for (int i = 0; i < N; ++i) { int X, Y; cin >> X >> Y; int dx = X - X0; int dy = Y - Y0; int d = gcd(dx, dy); dx /= d; dy /= d; if (dy < 0) { dy = -dy; dx = -dx; } if (dx == 0) dy = 1; S.insert(make_pair(dx, dy)); } cout << S.size() << "\n"; }
Codeforces Round #291 (Div. 2)
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原文地址：http://www.cnblogs.com/BigBallon/p/4293617.html
