BZOJ3858: Number Transformation

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=3858

题解：设第i个数为i*a;第i+1个数为（i+1）*b。则（i+1）*b>i*a;b>a-(a/(i+1));那么如果a<i+1;b的值就会不变

此时k*b就为所求；而由于a在不断的变小，a<i+1;在数值比较大的情况下就能实现
这么乱搞都行？！
代码：

#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<string>

#define inf 1000000000

#define maxn 200000+5

#define maxm 200000+5

#define eps 1e-10

#define ll long long

#define pa pair<int,int>

#define for0(i,n) for(int i=0;i<=(n);i++)

#define for1(i,n) for(int i=1;i<=(n);i++)

#define for2(i,x,y) for(int i=(x);i<=(y);i++)

#define for3(i,x,y) for(int i=(x);i>=(y);i--)

#define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)

#define for5(n,m) for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)

#define mod 1000000007

using namespace std;

inline int read()

{

    int x=0,f=1;char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}

    return x*f;

}

int main()

{

    freopen("input.txt","r",stdin);

    freopen("output.txt","w",stdout);
    ll n,m;int cs=0;

    while(scanf("%lld%lld",&n,&m)&&(n||m))
    {
        ll i,j;
        for(i=1;i<=m;i++)
        {
            j=n/i+(n%i>0);
            if(j<i+1)break;
            n=j*i;
        }
        if(i<=m)n=j*m;
        printf("Case #%d: %lld
",++cs,n);
    }

    return 0;

}