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:
View Code
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\n",++cs,n); 85 } 86 87 return 0; 88 89 }