• 51nod 1010 只包含因子2 3 5的数 && poj


    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.

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cmath>
     4 #include <vector>
     5 #include <cstring>
     6 #include <string>
     7 #include <algorithm>
     8 #include <string>
     9 #include <set>
    10 #include <functional>
    11 #include <numeric>
    12 #include <sstream>
    13 #include <stack>
    14 #include <map>
    15 #include <queue>
    16 //#pragma comment(linker, "/STACK:102400000,102400000")
    17 #define CL(arr, val)    memset(arr, val, sizeof(arr))
    18 
    19 #define ll long long
    20 #define inf 0x7f7f7f7f
    21 #define lc l,m,rt<<1
    22 #define rc m + 1,r,rt<<1|1
    23 #define pi acos(-1.0)
    24 
    25 #define L(x)    (x) << 1
    26 #define R(x)    (x) << 1 | 1
    27 #define MID(l, r)   (l + r) >> 1
    28 #define Min(x, y)   (x) < (y) ? (x) : (y)
    29 #define Max(x, y)   (x) < (y) ? (y) : (x)
    30 #define E(x)        (1 << (x))
    31 #define iabs(x)     (x) < 0 ? -(x) : (x)
    32 #define OUT(x)  printf("%I64d
    ", x)
    33 #define lowbit(x)   (x)&(-x)
    34 #define Read()  freopen("a.txt", "r", stdin)
    35 #define Write() freopen("b.txt", "w", stdout);
    36 #define maxn 1010
    37 #define maxv 1010
    38 #define mod 1000000000
    39 using namespace std;
    40 ll p[1555];
    41 int main()
    42 {
    43     //freopen("a.txt","r",stdin);
    44     int i=1,j=1,k=1;
    45     p[1]=1;
    46     for(int l=2;l<=1500;l++)
    47     {
    48         p[l]=min(p[i]*2,min(p[j]*3,p[k]*5));
    49         if(p[l]==p[i]*2) i++;
    50         if(p[l]==p[j]*3) j++;
    51         if(p[l]==p[k]*5) k++;
    52         //printf("%lld
    ",p[l]);
    53     }
    54     int n;
    55     while(~scanf("%d",&n)&&n)
    56     {
    57         printf("%lld
    ",p[n]);
    58     }
    59     return 0;
    60 }

    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.

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cmath>
     4 #include <vector>
     5 #include <cstring>
     6 #include <string>
     7 #include <algorithm>
     8 #include <string>
     9 #include <set>
    10 #include <functional>
    11 #include <numeric>
    12 #include <sstream>
    13 #include <stack>
    14 #include <map>
    15 #include <queue>
    16 //#pragma comment(linker, "/STACK:102400000,102400000")
    17 #define CL(arr, val)    memset(arr, val, sizeof(arr))
    18 
    19 #define ll long long
    20 #define inf 0x7f7f7f7f
    21 #define lc l,m,rt<<1
    22 #define rc m + 1,r,rt<<1|1
    23 #define pi acos(-1.0)
    24 
    25 #define L(x)    (x) << 1
    26 #define R(x)    (x) << 1 | 1
    27 #define MID(l, r)   (l + r) >> 1
    28 #define Min(x, y)   (x) < (y) ? (x) : (y)
    29 #define Max(x, y)   (x) < (y) ? (y) : (x)
    30 #define E(x)        (1 << (x))
    31 #define iabs(x)     (x) < 0 ? -(x) : (x)
    32 #define OUT(x)  printf("%I64d
    ", x)
    33 #define lowbit(x)   (x)&(-x)
    34 #define Read()  freopen("a.txt", "r", stdin)
    35 #define Write() freopen("b.txt", "w", stdout);
    36 #define maxn 1010
    37 #define maxv 1010
    38 #define mod 1000000000
    39 using namespace std;
    40 ll p[10999];
    41 
    42 ll erfen(ll n)
    43 {
    44     int l=1,r=10999;
    45     while(l<=r)
    46     {
    47         int mid=(l+r)/2;
    48         if(p[mid]==n) return p[mid];
    49         else if(p[mid]>n) r=mid-1;
    50         else l=mid+1;
    51     }
    52     if(p[l]<n) l++;
    53     return p[l];
    54 }
    55 int main()
    56 {
    57     freopen("a.txt","r",stdin);
    58     int i=1,j=1,k=1;
    59     p[1]=1;
    60     for(int l=2;l<=10999;l++)
    61     {
    62         p[l]=min(p[i]*2,min(p[j]*3,p[k]*5));
    63         if(p[l]==p[i]*2) i++;
    64         if(p[l]==p[j]*3) j++;
    65         if(p[l]==p[k]*5) k++;
    66         //printf("%lld
    ",p[l]);
    67     }
    68     //printf("%lld
    ",p[10000]);
    69     int t;
    70     ll n;
    71     scanf("%d",&t);
    72     while(t--)
    73     {
    74         scanf("%lld",&n);
    75         {
    76             if(n==1) printf("2
    ");
    77             else
    78             printf("%lld
    ",erfen(n));
    79         }
    80     }
    81     return 0;
    82 }
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  • 原文地址:https://www.cnblogs.com/nowandforever/p/4569165.html
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