推迟了15分钟开始，B卡住不会，最后弃疗，rating只涨一分。。。
水(暴力枚举) A - 2Char
/************************************************
* Author :Running_Time
* Created Time :2015/11/4 星期三 21:33:17
* File Name :A.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
char s[110][1010];
int len[110];
bool ok[110];
vector<char> vec[110];
int main(void) {
int n; scanf ("%d", &n);
for (int i=1; i<=n; ++i) {
scanf ("%s", s[i] + 1);
len[i] = strlen (s[i] + 1);
}
memset (ok, true, sizeof (ok));
int vis[30];
for (int i=1; i<=n; ++i) {
memset (vis, 0, sizeof (vis));
int tot = 0;
for (int j=1; j<=len[i]; ++j) {
if (!vis[s[i][j]-'a']) {
vec[i].push_back (s[i][j]);
vis[s[i][j]-'a'] = 1; tot++;
}
else continue;
if (tot > 2) {
ok[i] = false; break;
}
}
}
int ans = 0;
for (char a='a'; a<='z'; ++a) {
for (char b='a'; b<='z'; ++b) {
int tmp = 0;
for (int i=1; i<=n; ++i) {
if (!ok[i]) continue;
bool flag = true;
for (int j=0; j<vec[i].size (); ++j) {
char c = vec[i][j];
if (c != a && c != b) {
flag = false;
}
}
if (flag) tmp += len[i];
}
ans = max (ans, tmp);
}
}
printf ("%d
", ans);
//cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";
return 0;
}
sorting B - Anton and Lines
题意：给了一些直线，问是否在横坐标(x1, x2)范围内有相交的点
分析：很好想到每条直线与x = x1以及x = x2的直线的交点，那么满足相交的条件是y11 < y12 && y12 > y22，也就是逆序对。这样少掉了正好在x1或x2相交的情况，一种方法是L += EPS, R -= EPS，还有一种是排序。还有升级版的问题
/************************************************
* Author :Running_Time
* Created Time :2015/11/4 星期三 21:33:17
* File Name :B_2.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
struct Y {
double y1, y2;
bool operator < (const Y &r) const {
return y1 < r.y1;
}
}y[N];
/*
快速读入输出(读入输出外挂)!--黑科技
使用场合：huge input (1e6以上)
*/
inline int read(void) {
int f = 1, ret = 0; char ch = getchar ();
while ('0' > ch || ch > '9') {
if (ch == '-') f = -1;
ch = getchar ();
}
while ('0' <= ch && ch <= '9') {
ret = ret * 10 + ch - '0';
ch = getchar ();
}
return ret * f;
}
int main(void) {
int n; scanf ("%d", &n);
double x1, x2; scanf ("%lf%lf", &x1, &x2);
x1 += EPS; x2 -= EPS;
double k, b;
for (int i=1; i<=n; ++i) {
scanf ("%lf%lf", &k, &b);
y[i].y1 = k * x1 + b;
y[i].y2 = k * x2 + b;
}
sort (y+1, y+1+n);
bool flag = false;
for (int i=2; i<=n; ++i) {
if (y[i].y2 < y[i-1].y2) {
flag = true; break;
}
}
if (flag) puts ("YES");
else puts ("NO");
// cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";
return 0;
}
/************************************************
* Author :Running_Time
* Created Time :2015/11/4 星期三 21:33:17
* File Name :B.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
struct Y {
ll y1, y2;
bool operator < (const Y &r) const {
return y1 < r.y1 || (y1 == r.y1 && y2 < r.y2);
}
}y[N];
/*
快速读入输出(读入输出外挂)!--黑科技
使用场合：huge input (1e6以上)
*/
inline int read(void) {
int f = 1, ret = 0; char ch = getchar ();
while ('0' > ch || ch > '9') {
if (ch == '-') f = -1;
ch = getchar ();
}
while ('0' <= ch && ch <= '9') {
ret = ret * 10 + ch - '0';
ch = getchar ();
}
return ret * f;
}
int main(void) {
int n; scanf ("%d", &n);
int x1, x2; scanf ("%d%d", &x1, &x2);
ll k, b;
for (int i=1; i<=n; ++i) {
k = read (); b = read ();
y[i].y1 = k * x1 + b;
y[i].y2 = k * x2 + b;
}
sort (y+1, y+1+n);
bool flag = false;
for (int i=2; i<=n; ++i) {
if (y[i].y2 < y[i-1].y2) {
flag = true; break;
}
}
if (flag) puts ("YES");
else puts ("NO");
// cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";
return 0;
}