【1-n】区间覆盖 TOJ4168+BZOJ1192

Xiao Ming is very interesting for array. He given a sorted positive integer array and an integer n.
We need add elements to the array such that the sum of subset of array cover the range [1, n]. You must return the minimum number of add elements.
Xiao Ming worry about you can't understand this problem. Now, he gives you an example:
There is an array [1, 3], n = 6, the answer is 1. Because we only need to add 2 into array. After add operator, the array is [1,2,3]. The subsets are: [1], [2], [3], [1,2], [1,3], [2,3], [1,2,3]. Possible sums are 1, 2, 3, 4, 5, 6, which now covers the range [1, 6].

Input

The input consists of multiple test cases. The first line contains an integer

Output

Return the minimum number of add elements.

Sample Input

1
3 20
1 5 10

Sample Output

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long num[100861];
int main()
{
    int T;
    for(scanf("%d",&T);T--;){
        long long n,m;
        scanf("%lld%lld",&n,&m);
        for(int i=1;i<=n;++i) scanf("%lld",num+i);
        sort(num+1,num+1+n);
        long long sum=0;
        int ct=0,pos=1;
        for(long long i=1;i<=m;++i)
        {
            if(num[pos]<=i&&pos<=n) {
                sum+=num[pos];
                ++pos;
                i=sum;
            }
            else {
                ++ct;
                sum+=i;
                i=sum;
            }
        }
        printf("%d
",ct);
    }
}

1192: [HNOI2006]鬼谷子的钱袋

Time Limit: 10 Sec Memory Limit: 162 MB
Submit: 3752 Solved: 2705
[Submit][Status][Discuss]

Description

鬼谷子非常聪明，正因为这样，他非常繁忙，经常有各诸侯车的特派员前来向他咨询时政。有一天，他在咸阳游历的时候，朋友告诉他在咸阳最大的拍卖行（聚宝商行）将要举行一场拍卖会，其中有一件宝物引起了他极大的兴趣，那就是无字天书。但是，他的行程安排得很满，他他已经买好了去邯郸的长途马车标，不巧的是出发时间是在拍卖会快要结束的时候。于是，他决定事先做好准备，将自己的金币数好并用一个个的小钱袋装好，以便在他现有金币的支付能力下，任何数目的金币他都能用这些封闭好的小钱的组合来付账。鬼谷子也是一个非常节俭的人，他想方设法使自己在满足上述要求的前提下，所用的钱袋数最少，并且不有两个钱袋装有相同的大于1的金币数。假设他有m个金币，你能猜到他会用多少个钱袋，并且每个钱袋装多少个金币吗？

Input

包含一个整数，表示鬼谷子现有的总的金币数目m。其中，1≤m ≤1000000000。

Output

只有一个整数h，表示所用钱袋个数

Sample Input

Sample Output

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int main()
{
    long long sum=0,m,ans=1,ct,tt=0;
    scanf("%lld",&m);
    while(1)
    {
        if(sum+ans>m) break;
        sum+=ans;
        ans<<=1;
        ++tt;
    }
    ct=m-sum;
    if(!ct) printf("%lld ",tt);
    else printf("%lld ",tt+1);
}