Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2^wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there‘s no more weight left. Malek asked her to minimize the number of steps.
Duff is a competitive programming fan. That‘s why in each step, she can only lift and throw away a sequence of weights 2^a1, ..., 2^ak if and only if there exists a non-negative integer x such that 2^a1 + 2^a2 + ... + 2^ak = 2^x, i. e. the sum of those numbers is a power of two.
Duff is a competitive programming fan, but not a programmer. That‘s why she asked for your help. Help her minimize the number of steps.
The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.
The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.
Print the minimum number of steps in a single line.
5
1 1 2 3 3
2
4
0 1 2 3
4
In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it‘s not possible to do it in one step because their sum is not a power of two.
In the second sample case: The only optimal way is to throw away one weight in each step. It‘s not possible to do it in less than 4 steps because there‘s no subset of weights with more than one weight and sum equal to a power of two.
题意：给定n个数w1, w2, w3,.......wn, 然后从这个n数中找出这样的一个子序列a1,a2,a3....ak 使得 2^a1+2^a2.....+2^ak = 2^x, 然后可以删除这个序列，
那么最少经过几步可以全部将这n个数删除！
思路：其实就是 二进制数 相加的过程，n个数对应n个二进制数，从最低位到最高位相加，得到最后的二进制数中 1 的个数就是答案！
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<map> using namespace std; map<int, int, less<int> >mp; int main(){ int n; scanf("%d", &n); while(n--){ int x; scanf("%d", &x); mp[x]++; } int ans = 0; for(map<int,int>::iterator it = mp.begin(); it!=mp.end(); ++it){ if(it->second>1) mp[it->first + 1] += it->second / 2; if(it->second%2 != 0) ++ans; } printf("%d\n", ans); return 0; }
Codeforces Round #326 (Div. 2) C. Duff and Weight Lifting
原文：http://www.cnblogs.com/hujunzheng/p/4895835.html