CodeForces Gym100935 水题专场
CodeForces Gym100935
比赛链接： CodeForces Gym100935
A Time
水题~
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define fst first
#define snd second
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef __int64 LL;
//typedef long long LL;
typedef unsigned int uint;
typedef pair PII;
typedef pair PLL;
const double PI = 3.1415926536;
const double eps = 1e-6;
const int MAXN = 300 + 5;
int T;
int H, M, S;
int h, m, s;
int A, B;
int main() {
#ifndef ONLINE_JUDGE
FIN;
// FOUT;
#endif // ONLINE_JUDGE
int cas = 0;
scanf ("%d", &T);
while (T --) {
scanf ("%d %d %d", &H, &M, &S);
scanf ("%d %d %d", &h, &m, &s);
A = H * 3600 + M * 60 + S;
B = h * 3600 + m * 60 + s;
printf ("Case %d: %s\n", ++cas, A == B ? "Yes" : "No");
}
return 0;
} B. Weird Cryptography
题意描述：给定N个单词，你可以把他们组成若干个集合，每个集合不能有相同的单词。然后呢，取第i个集合的大小比如为|Si|，|Si| <= 9 ，让所有集合的大小组成一个十进制数字。让这个数字最小， 求出这个数。
解题思路：首先， 要让数字最小， 就必须让这个数位数最少。然后就是肯定是要让最大的那个集合尽可能的大，最小的那个集合尽可能的小。
先用map 对每个字符串标记一下。然后cnt数组记录一下每个元素出现的次数。然后，遍历每个cnt，对应打好一个标记，最好前缀和搞一下。
#include
#include C. OCR
字符串水题
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define fst first
#define snd second
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef __int64 LL;
//typedef long long LL;
typedef unsigned int uint;
typedef pair PII;
typedef pair PLL;
const double PI = 3.1415926536;
const double eps = 1e-6;
const int MAXN = 300 + 5;
int T;
int N, M;
char G[MAXN][MAXN];
int main() {
#ifndef ONLINE_JUDGE
FIN;
// FOUT;
#endif // ONLINE_JUDGE
int cas = 0;
scanf ("%d", &T);
while (T--) {
scanf ("%d %d", &N, &M);
for (int i = 0; i < N; i++) {
scanf ("%s", G[i]);
}
int x1, y1, x2, y2;
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (G[i][j] == '*') {
x1 = i;
y1 = j;
goto F1;
}
}
}
F1:
;
for (int i = N - 1; i >= 0; i--) {
for (int j = M - 1; j >= 0; j--) {
if (G[i][j] == '*') {
x2 = i;
y2 = j;
goto F2;
}
}
}
F2:
;
int res = 0;
for (int i = x1 + 1; i < x2; i++) {
if (G[i][y1 + 1] == '*') res = 1;
}
if (res) printf ("Case %d: Eight\n", ++cas);
else printf ("Case %d: Zero\n", ++cas);
}
return 0;
} D. Enormous Carpet
裸的快速幂#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define fst first
#define snd second
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef __int64 LL;
//typedef long long LL;
typedef unsigned int uint;
typedef pair PII;
typedef pair PLL;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int MAXN = 100 + 5;
LL N, K, A, C;
LL P[MAXN];
LL quick_pow (LL a, LL b, LL mod) {
LL ret = 1LL;
while (b) {
if (b & 1LL) ret = ret * a % mod;
a = a * a % mod;
b >>= 1LL;
}
return ret;
}
int main() {
#ifndef ONLINE_JUDGE
FIN;
// FOUT;
#endif // ONLINE_JUDGE
LL cas = 0LL;
while (~scanf ("%I64d %I64d %I64d %I64d", &N, &K, &A, &C) && N) {
printf ("Case %I64d:\n", ++cas);
for (LL i = 0; i < C; i++) {
scanf ("%I64d", &P[i]);
// cout << quick_pow (K, N, INF) * A << endl;
printf ("%I64d%c", A % P[i] * quick_pow (K, N, P[i]) % P[i], i == C - 1 ? '\n' : ' ');
}
}
return 0;
} E. Pairs
题意描述： 有5个数字(X1, X2, X3, X4, X5) ，他们两两组成可以组成10个数对。现在一直这每个数对中两个数之和，让你还原这5个数字。
解题思路：水题~枚举。
首先对这个10数字进行排序，然后
已经确定的是有(X1, X2) (X1, X3), (X3,X5), (X4,X5) 这四个数对 对应的和是多少。
然后只需要枚举 (X2, X3) 和 (X3, X4) 这两个数对的和就Ok了。
X[1] = (X1 + X2) + (X1 + X3) - (X2 + X3)
其他的同理。
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define fst first
#define snd second
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef __int64 LL;
//typedef long long LL;
typedef unsigned int uint;
typedef pair PII;
typedef pair PLL;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int MAXN = 10 + 5;
int T, cas;
int S[MAXN];
int X[MAXN];
int G[MAXN][MAXN];
int F[MAXN];
bool check() {
// for (int i = 1; i <= 5; i++) {
// if (X[i] <= 0) return false;
// }
// for (int i = 2; i <= 5; i++) {
// if (X[i - 1] > X[i]) return false;
// }
int k = 0;
for (int i = 1; i <= 5; i++) {
for (int j = i + 1; j <= 5 ; j++) {
F[++ k] = X[i] + X[j];
}
}
sort (F + 1, F + 10 + 1);
for (int i = 1; i <= 10; i++) {
if (S[i] != F[i]) return false;
}
return true;
}
int main() {
#ifndef ONLINE_JUDGE
FIN;
// FOUT;
#endif // ONLINE_JUDGE
scanf ("%d", &T);
while (T --) {
for (int i = 1; i <= 10; i ++) {
scanf ("%d", &S[i]);
}
sort (S + 1, S + 10 + 1);
G[1][2] = S[1];
G[1][3] = S[2];
G[3][5] = S[9];
G[4][5] = S[10];
bool suc = false;
for (int i = 3; i <= 8; i++) {
if (suc) break;
G[2][3] = S[i];
for (int j = 8; j > i; j--) {
G[3][4] = S[j];
X[1] = G[1][2] + G[1][3] - G[2][3];
if (X[1] & 1) continue;
X[1] >>= 1;
X[5] = G[3][5] + G[4][5] - G[3][4];
if (X[5] & 1) continue;
X[5] >>= 1;
X[2] = G[1][2] - X[1];
X[3] = G[2][3] - X[2];
X[4] = G[4][5] - X[5];
if (check() ) {
suc = true;
break;
}
if (suc) break;
}
}
sort(X + 1, X + 5 + 1);
printf ("Case %d: %d %d %d %d %d\n", ++ cas, X[1], X[2], X[3], X[4], X[5]);
}
return 0;
}
F. A Poet Computer
题意描述：给定N个单词，这些单词中，如果有三个或者三个以上单词尾部相同，则就定义为"Common Suffix"， 求"Common Suffix"最长的长度，以及出现的次数。
解题思路： 很显然，直接将单词逆序加入到字典树中，然后直接统计就好了。
解题思路： 很显然，直接将单词逆序加入到字典树中，然后直接统计就好了。
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define fst first
#define snd second
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef __int64 LL;
//typedef long long LL;
typedef unsigned int uint;
typedef pair PII;
typedef pair PLL;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int MAXN = 100 * 1000 + 5;
int T, cas, N;
struct Node {
int cnt;
Node* pNext[26];
} *pRoot;
Node Mem[MAXN];
int SIZE;
int MaxLen, Times;
Node* new_node() {
Node* p = &Mem[SIZE ++];
p->cnt = 0;
for (int i = 0; i < 26; i++) p->pNext[i] = NULL;
return p;
}
void init_trie() {
SIZE = 0;
pRoot = new_node();
}
void add_word (char s[], int len) {
Node* p = pRoot;
int pos;
for (int i = len - 1, k = 1; i >= 0; i--, k++) {
pos = s[i] - 'a';
if (p->pNext[pos] == NULL) {
p->pNext[pos] = new_node();
}
p->pNext[pos]->cnt ++;
p = p->pNext[pos];
if (p->cnt >= 3) {
if (MaxLen < k) {
MaxLen = k;
Times = p->cnt;
} else if (MaxLen == k) {
Times = max (Times, p->cnt);
}
}
}
}
char buf[110];
int main() {
#ifndef ONLINE_JUDGE
FIN;
// FOUT;
#endif // ONLINE_JUDGE
scanf ("%d", &T);
while (T--) {
init_trie();
MaxLen = Times = 0;
scanf ("%d", &N);
for (int i = 0; i < N; i++) {
scanf ("%s", buf);
for (int i = 0; i < strlen (buf); i++) {
if (buf[i] >= 'A' && buf[i] <= 'Z') buf[i] = buf[i] - 'A' + 'a';
}
add_word (buf, strlen (buf) );
}
printf ("Case %d:\n", ++cas);
if (N < 3) {
printf ("0 0");
continue;
}
printf ("%d %d\n", MaxLen, Times);
}
return 0;
} H. Bend Test
题目描述：有P部手机需要进行压力测试（每部手机的耐压值是一样的），压力值从1~M，(P∈[1, 50], M ∈[1, 1000])， 当压力值大于手机的耐压值时，这部手机就被摧毁了，不能继续进行测试了， 现在要你求出，在最坏的情况下，最少需要多少次可以测出压力值。
解题思路：裸DP。
用dp[p][m] 表示 p部手机 最大压力值为m 的答案。
然后有，
dp[p][m] = min (dp[p][m], (dp[p - 1][x - 1], dp[p][m - x]) + 1);
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define fst first
#define snd second
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef __int64 LL;
//typedef long long LL;
typedef unsigned int uint;
typedef pair PII;
typedef pair PLL;
const int INF = 0x3f3f3f3f;
const int MAXP = 50 + 5;
const int MAXM = 1000 + 5;
int T, cas;
int M, P;
int dp[MAXP][MAXM];
// 求log2 的 值
int F (int x) {
int ret = 0, y = 1;
while (y < x) y <<= 1, ret++;
return ret;
}
int main() {
#ifndef ONLINE_JUDGE
FIN;
#endif // ONLINE_JUDGE
for (int p = 1; p < MAXP; p++) dp[p][1] = 1, dp[p][p] = F (p + 1);
for (int m = 1; m < MAXM; m++) dp[1][m] = m;
for (int p = 2, t; p < MAXP; p++) {
for (int m = 2; m <= p; m++) dp[p][m] = dp[m][m];
for (int m = p + 1; m < MAXM; m++) {
dp[p][m] = INF;
for (int x = 2; x < m; x++) {
t = max (dp[p - 1][x - 1], dp[p][m - x]);
dp[p][m] = min (dp[p][m], t + 1);
}
}
}
scanf ("%d", &T);
while (T --) {
scanf ("%d %d", &P, &M);
printf ("Case %d: %d\n", ++cas, dp[P][M]);
}
return 0;
} I. Farm
题目描述： 求矩形与圆相交的面积。已知矩形长宽>圆的半径，矩形左上角在圆中。
解题思路：刚开始没有好好看题，没看到
矩形左上角在圆中 结果就想 要根据矩形的位置分情况考虑，Too Young Too Simple!...
如图， 将矩形和圆的公共部分分割成 两个部分， 一个三角形，和一个弓形( 弦与弧围成的区域 )区域。分别求面积即可。
#include
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define fst first
#define snd second
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define fcout(x) cout << fixed << setprecision(6)
typedef __int64 LL;
typedef pair PII;
const int MAXN = 300 + 5;
int T, N, M;
double XC, YC, R;
double XB, YB, XU, YU;
double X1, X2, Y1, Y2;
double S1, S2, S3;
double alfa, dist;
int main() {
#ifndef ONLINE_JUDGE
FIN;
// FOUT;
#endif // ONLINE_JUDGE
int cas = 0;
scanf ("%d", &T);
while (T --) {
scanf ("%lf %lf %lf", &XC, &YC, &R);
scanf ("%lf %lf %lf %lf", &XB, &YB, &XU, &YU);
XU -= XC;
YU -= YC;
XB -= XC;
YB -= YC;
Y1 = YU;
X1 = sqrt (R * R - Y1 * Y1);
X2 = XB;
Y2 = -sqrt (R * R - X2 * X2);
dist = sqrt ( (X1 - X2) * (X1 - X2) + (Y1 - Y2) * (Y1 - Y2) );
alfa = acos ( (R * R * 2.0 - dist * dist) / (2.0 * R * R) );
S1 = 0.5 * fabs (X1 - X2) * fabs (Y1 - Y2);
S2 = (R * R * 0.5) * (alfa - sin (alfa) );
S3 = S1 + S2;
cout << "Case " << ++cas << ": ";
cout << fixed << setprecision(5) << S3 << endl;
}
return 0;
}