Codeforces 588C Duff and Weight Lifting 【数学】

题目链接：Codeforces 588C Duff and Weight Lifting

C. Duff and Weight Lifting
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 2wi 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 2a1, …, 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + … + 2ak = 2x, 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.

Input
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.

Output
Print the minimum number of steps in a single line.

Examples
input
5
1 1 2 3 3
output
2
input
4
0 1 2 3
output
4
Note
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：有n个数a[i]代表$2^{a[i]}$，每次可以去掉若干个数当它们和为2的幂次时。问最少需要多少次。

思路：一直合并就可以了，因为当a!=b时，$2^a + 2^b$肯定不是2的幂次。

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e7 + 10;
int a[MAXN];
int vis[MAXN];
int main()
{
    int n;
    while(scanf("%d", &n) != EOF) {
        CLR(vis, 0); int Max = 0;
        for(int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
            vis[a[i]]++;
            Max = max(Max, a[i]);
        }
        for(int i = 0; i <= Max; i++) {
            if(vis[i] % 2 == 0) {
                vis[i+1] += vis[i] / 2;
                vis[i] = 0;
            }
            else {
                vis[i+1] += vis[i] / 2;
                vis[i] = 1;
            }
        }
        Max++;
        while(1) {
            if(vis[Max] == 1 || vis[Max] == 0) {
                break;
            }
            if(vis[Max] & 1) {
                vis[Max+1] += vis[Max] / 2;
                vis[Max] = 1;
            }
            else {
                vis[Max+1] += vis[Max] / 2;
                vis[Max] = 0;
            }
            Max++;
        }
        int ans = 0;
        for(int i = 0; i <= Max; i++) {
            ans += vis[i];
        }
        printf("%d\n", ans);
    }
    return 0;
}