BestCoder Round #7

A(hdu4985) - Little Pony and Permutation

 1 #include <cstdio>
 2
 3 using namespace std;
 4
 5 #define NN    100010
 6 int g[NN], f[NN];
 7
 8 int main(void)
 9 {
10     int N;
11     while(scanf("%d", &N) > 0) {
12         for(int i=1; i<=N; ++i) scanf("%d", &f[i]);
13         for(int i=1; i<=N; ++i) g[i] = 0;
14         for(int i=1; i<=N; ++i) {
15             if (!g[i]) {
16                 putchar(‘(‘);
17                 int t = i;
18                 do {
19                     printf(t != i? " %d" : "%d", t);
20                     g[t] = 1;
21                     t = f[t];
22                 } while(t != i);
23                 putchar(‘)‘);
24             }
25         }
26         putchar(‘\n‘);
27     }
28     return 0;
29 }

B(hdu4986) - Little Pony and Alohomora Part I

 1 #include <cstdio>
 2 #include <cmath>
 3 using namespace std;
 4
 5 #define NN    110000
 6 double g[NN + 1];
 7
 8 int main(void)
 9 {
10     for(int i=1; i<=NN; ++i) g[i] = g[i-1] + 1/(double)i;
11     int N;
12     while(scanf("%d", &N) > 0)
13         printf("%.4f\n", N<=NN ? g[N] : log(N+1) + 0.577215664);
14     return 0;
15 }
时间: 2024-10-13 00:52:18

BestCoder Round #7的相关文章

BestCoder Round #91

传送门:http://acm.hdu.edu.cn/search.php?field=problem&key=BestCoder+Round+%2391&source=1&searchmode=source A题:给你n种字母,每种字母有个权值vali,共cnti个,现在让你在里面挑出任意数量的字符,组合成一个字符串,该字符串的权值的计算方式为val1*1+val2*2+--+valn*n,让你输出字符串最大的权值是多少.这题很容易会有一个错误的贪心,就是把val为负的舍去,然后v

BestCoder Round #65 (ZYB&#39;s Game)

ZYB's Game Accepts: 672 Submissions: 1207 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Problem Description ZYBZYBZYB played a game named NumberBombNumber BombNumberBomb with his classmates in hiking:a host keeps a

hdu5418 BestCoder Round #52 (div.2) Victor and World ( floyd+状压dp)

Problem Description After trying hard for many years, Victor has finally received a pilot license. To have a celebration, he intends to buy himself an airplane and fly around the world. There are n countries on the earth, which are numbered from 1 to

hdu 5163 Taking Bus (BestCoder Round #27)

Taking Bus                                                               Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 501    Accepted Submission(s): 203 Problem Description Bestland has a v

BestCoder Round #11 (Div. 2) 前三题题解

题目链接: huangjing hdu5054 Alice and Bob 思路: 就是(x,y)在两个參考系中的表示演全然一样.那么仅仅可能在这个矩形的中点.. 题目: Alice and Bob Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 216    Accepted Submission(s): 166 Problem De

BestCoder Round #16

BestCoder Round #16 题目链接 这场挫掉了,3挂2,都是很sb的错误 23333 QAQ A:每个数字,左边个数乘上右边个数,就是可以组成的区间个数,然后乘的过程注意取模不然会爆掉 B:dp,dp[i][2]记录下第一长的LIS,和第二长的LIS,哎,转移的时候一个地方写挫掉了导致悲剧啊QAQ C:首先如果知道Nim游戏的,就很容易转化问题为,一些数字是否能选几个能否异或和为0,那么就是每个数字拆成40位,然后每一位异或和为0,这样就可以构造出40个方程,然后高斯消元求解,如果

BestCoder Round #11 (Div. 2) 题解

HDOJ5054 Alice and Bob Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 302    Accepted Submission(s): 229 Problem Description Bob and Alice got separated in the Square, they agreed that if they

BestCoder Round #9

BestCoder Round #9 题目链接 A:暴力枚举一个数字,就能求出另一个数字,for一遍即可 B:博弈,判断前n - 1个开头连续1的奇偶性即可 C:先预处理出每个点对应哪几个点,每次查询计算一次即可 代码: A: #include <cstdio> #include <cstring> #include <vector> #include <set> #include <algorithm> #include <cmath&g

hdu 4956 Poor Hanamichi BestCoder Round #5(数学题)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4956 Poor Hanamichi Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7    Accepted Submission(s): 4 Problem Description Hanamichi is taking part in

BestCoder Round #4 前两题 hdu 4931 4932

第一题太水了.. 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 int a[6]; 7 int main(){ 8 int cas; 9 scanf( "%d", &cas ); 10 while( cas-- ){ 11 for( int i = 0; i <