51nod 1010 只包含因子2 3 5的数 && poj - 1338 Ugly Numbers(打表)

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1010

http://poj.org/problem?id=1338

首先poj这题要找出第n个丑数,想到要打表,因为每一个丑数都是由前一个丑数*2或者*3或者*5得到,每次取3种结果种较小的那一个放入数组中.

打表的时候要注意保持数组有序.

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>
 //#pragma comment(linker, "/STACK:102400000,102400000")
 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1010
 #define maxv 1010
 #define mod 1000000000
 using namespace std;
 ll p[];
 int main()
 {
     //freopen("a.txt","r",stdin);
     int i=,j=,k=;
     p[]=;
     for(int l=;l<=;l++)
     {
         p[l]=min(p[i]*,min(p[j]*,p[k]*));
         if(p[l]==p[i]*) i++;
         if(p[l]==p[j]*) j++;
         if(p[l]==p[k]*) k++;
         //printf("%lld\n",p[l]);
     }
     int n;
     while(~scanf("%d",&n)&&n)
     {
         printf("%lld\n",p[n]);
     }
     return ;
 }

51nod这题让你求>=n的丑数,因为n很大,所以打完表后需要二分查找,打表的时候要注意直到大于10^18,上限可以自己试几次.

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>
 //#pragma comment(linker, "/STACK:102400000,102400000")
 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1010
 #define maxv 1010
 #define mod 1000000000
 using namespace std;
 ll p[];

 ll erfen(ll n)
 {
     int l=,r=;
     while(l<=r)
     {
         int mid=(l+r)/;
         if(p[mid]==n) return p[mid];
         else if(p[mid]>n) r=mid-;
         else l=mid+;
     }
     if(p[l]<n) l++;
     return p[l];
 }
 int main()
 {
     freopen("a.txt","r",stdin);
     int i=,j=,k=;
     p[]=;
     for(int l=;l<=;l++)
     {
         p[l]=min(p[i]*,min(p[j]*,p[k]*));
         if(p[l]==p[i]*) i++;
         if(p[l]==p[j]*) j++;
         if(p[l]==p[k]*) k++;
         //printf("%lld\n",p[l]);
     }
     //printf("%lld\n",p[10000]);
     int t;
     ll n;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%lld",&n);
         {
             if(n==) printf("2\n");
             else
             printf("%lld\n",erfen(n));
         }
     }
     return ;
 }