http://acm.hdu.edu.cn/showproblem.php?pid=2051 Problem Description Give you a number on base ten,you should output it on base two.(0 < n < 1000)   Input For each case there is a postive number n on base ten, end of file.   Output For each case outpu…

问题: 之前做过类似题,但这次仍然不能解决相关问题. 字符串倒过来输:StringBuffer str=new StringBuffer(s); s=str.reverse().toString() Bitset Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 15548    Accepted Submission(s): 11804…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=5313 题意: 给出n个顶点,m条边,问最多添加多少条边使之构成一个完全二分图 存储结构: bitset     [用法详情:http://blog.csdn.net/piaocoder/article/details/47177891] 用时:624ms 思路: 二分图的总边数即:n*m(假设一个有n个点,另一个有m个点) 题目是给出总共的点数为n,间接求最大的边数 想到一个小学题:给出长度为n的绳子,…

题目大意: 添加尽可能少的边,最后使图形成二分图 一开始将图区分成一个个联通分量,根据二分图染色,计算出每个联通分量的黑色点和白色点的个数 希望添加的边最少,那么合并的时候,希望黑白块尽可能平均,这无疑背包dp做,但超时了...T T 跟着题解说的bitset,学了一下,果然总共10000个点不到,那么只要用bitset上的某一位代表取到的值即可- -,好神奇..这里用的是或运算 #include <cstdio> #include <cstring> #include <i…

Bitset Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14685    Accepted Submission(s): 11173 Problem Description Give you a number on base ten,you should output it on base two.(0 < n < 1000)  …

Rikka with Candies Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1452    Accepted Submission(s): 634 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situatio…

ps:这道题是题目坑爹了...题目说不考虑n=0的...但其实要考虑...醉了 中文意思:输入一个10进制的数,输出他的二进制数 代码: #include "stdio.h"int main(){ int n,i,a[1000],num;    while(~scanf("%d",&n)){     num=0;     if(n==0){      printf("0\n");continue;     }     if(n==1){ …

题意是将十进制数转换成二进制数. 从网上找到的十进制转 k 进制的做法,代码如下: #include <bits/stdc++.h> using namespace std; ','A','B','C','D','E','F'}; void conv(int n,int k)//k 为进制 { int r; r = n%k; n = n/k; if(n) conv(n,k); printf("%c",a[r]); } int main() { int n,k; while(…

https://vjudge.net/contest/262753#overview C - Regular Number HDU - 5972 bitset temp, temp[i]=1表示 此前i个位置都能完全匹配,&=bt[x-'0']来递推 int n,a[1111][12]; char s[5000050]; int main() { while(~scanf("%d",&n)) { bitset<1000> bt[11]; re(i,0,9)b…

2047  阿牛的EOF牛肉串 #include<stdio.h> int main(){ int n,i; _int64 s[]; while(~scanf("%d",&n)){ s[]=;s[]=; ;i<=n;i++){ s[i] = s[i-]* + s[i-]*; } printf("%I64d\n",s[n]); } } 2048  神.上帝以及老天爷 #include<stdio.h> int main(){ in…