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：

 1 #include<cstdio>
 2 
 3 #include<cstdlib>
 4 
 5 #include<cmath>
 6 
 7 #include<cstring>
 8 
 9 #include<algorithm>
10 
11 #include<iostream>
12 
13 #include<vector>
14 
15 #include<map>
16 
17 #include<set>
18 
19 #include<queue>
20 
21 #include<string>
22 
23 #define inf 1000000000
24 
25 #define maxn 200000+5
26 
27 #define maxm 200000+5
28 
29 #define eps 1e-10
30 
31 #define ll long long
32 
33 #define pa pair<int,int>
34 
35 #define for0(i,n) for(int i=0;i<=(n);i++)
36 
37 #define for1(i,n) for(int i=1;i<=(n);i++)
38 
39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
40 
41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
42 
43 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
44 
45 #define for5(n,m) for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
46 
47 #define mod 1000000007
48 
49 using namespace std;
50 
51 inline int read()
52 
53 {
54 
55     int x=0,f=1;char ch=getchar();
56 
57     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
58 
59     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
60 
61     return x*f;
62 
63 }
64 
65 int main()
66 
67 {
68 
69     freopen("input.txt","r",stdin);
70 
71     freopen("output.txt","w",stdout);
72     ll n,m;int cs=0;
73 
74     while(scanf("%lld%lld",&n,&m)&&(n||m))
75     {
76         ll i,j;
77         for(i=1;i<=m;i++)
78         {
79             j=n/i+(n%i>0);
80             if(j<i+1)break;
81             n=j*i;
82         }
83         if(i<=m)n=j*m;
84         printf("Case #%d: %lld
",++cs,n);
85     }
86 
87     return 0;
88 
89 }

View Code