[POJ3252]Round Numbers 试题描述 The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors', 'Ro, Sham, Bo', and a host of other names) in order to make arbitrary decisions s…

题目问区间有多少个数字的二进制0的个数大于等于1的个数. 用数学方法求出0到n区间的合法个数,然后用类似数位DP的统计思想. 我大概是这么求的,确定前缀的0和1,然后后面就是若干个0和若干个1的不重复全排列数.. 写得挺痛苦的..另外A[i][j]表示i个0和j个1的不重复全排列数,即A[i][j]=(i+j)!/i!/j!,这个可以从A[i-1][j]或A[i][j-1]求出来,这样就不用担心乘法溢出了. #include<cstdio> #include<cstring> us…

Round Numbers Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7625   Accepted: 2625 Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors',…

题目链接:http://poj.org/problem?id=3252 Round Numbers Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14640   Accepted: 5881 Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (al…

题目传送门 Round Numbers Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 16439   Accepted: 6776 Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scis…

Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors', 'Ro, Sham, Bo', and a host of other names) in order to make arbitrary decisions such as who gets…

题目链接 POJ3252 题解 为什么每次写出数位dp都如此兴奋? 因为数位dp太苟了 因为我太弱了 设\(f[i][0|1][cnt1][cnt0]\)表示到二进制第\(i\)位,之前是否达到上界,前面已经有\(cnt1\)个\(1\),\(cnt0\)个\(0\)时的方案数 显然当\(cnt1 = 0\)时就不存在任何前导数字了 然后就记忆化搜索 分类讨论各种转移 [为什么我写得好麻烦QAQ是不是我姿势不对] #include<iostream> #include<cstdio>…

地址 拆成2进制位做dp记搜就行了,带一下前导0,将0和1的个数带到状态里面,每种0和1的个数讨论一下,累加即可. WA记录:line29. #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #define dbg(x) cerr<<#x<<" = "<<x<…

题目大意: 求区间 \([x,y]\) 范围内有多少数的二进制表示中的'0'的个数 \(\ge\) '1'的个数. 解题思路: 使用 数位DP 解决这个问题. 我们设状态 f[pos][num0][num1][all0] 表示在: 当前所在数位为 pos : 当前选择的'0'的个数为 num0: 当前选择的'1'的个数为 num1: 到当前位位置是不是前面的数都是前导零(如果都是前导0则 all0==true,否则 all==false). 下的方案数. 我们开函数 dfs(int pos, i…

Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors', 'Ro, Sham, Bo', and a host of other names) in order to make arbitrary decisions such as who gets…