51Nod 1596 搬货物

思路：题目要求每次搬的货物的重量和必须是2的次方数，给出得数据也是2的次幂，显然，两个货物的重量和为2的次方数，当且仅当这两个货物的次幂相等，所以我们始终保持数组是升序的，当w[i]==w[i+1]时，去掉w[i],w[i+1]++后保留，否则去掉w[i],保留w[i+1]，方案数+1。可以用优先队列优化。
但这题卡了输入，需要使用输入挂。。。。。。。

#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cstring>#include <memory>#include <cctype>#include <bitset>#include <string>#include <vector>#include <climits>#include <cstring>#include <iostream>#include <iomanip>#include <algorithm>#include <functional>//#define FIN freopen("input.txt","r",stdin);//#define FOUT freopen("output.txt","w+",stdout);using namespace std;typedef long long ll;const int INF = 0x3f3f3f3f;const int mod = 1e9 + 7;const double eps=1e-8;const double Pi=acos(-1.0);const int N=1000100;template <class T>//适用于正整数的输入挂inline void scan_d(T &ret){    char c;    ret = 0;    while ((c = getchar()) < '0' || c > '9');    while (c >= '0' && c <= '9')    {        ret = ret * 10 + (c - '0'), c = getchar();    }}struct cmp{     bool operator ()(int x, int y)    {        return x > y;//小的优先级高    }};priority_queue<int,vector<int>,cmp> que;int main(){    int n;    scanf("%d",&n);    int num;    for(int i=0;i<n;i++)    {        scan_d(num);        que.push(num);    }    int ans=0;    int tmp1,tmp2;    while(!que.empty())    {        if(int(que.size()==1))        {            ans++;            break;        }        tmp1=que.top();        que.pop();        tmp2=que.top();        que.pop();        if(tmp1==tmp2)        {            tmp2++;            que.push(tmp2);        }        else        {            ans++;            que.push(tmp2);        }    }    printf("%d\n",ans);}