• hdu_5104 Primes Problem()


    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5104

    rimes Problem

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2844    Accepted Submission(s): 1277


    Problem Description
    Given a number n, please count how many tuple(p1, p2, p3) satisfied that p1<=p2<=p3, p1,p2,p3 are primes and p1 + p2 + p3 = n.
     
    Input
    Multiple test cases(less than 100), for each test case, the only line indicates the positive integer n(n10000).
     
    Output
    For each test case, print the number of ways.
     
    Sample Input
    3 9
     
    Sample Output
    0 2
     
    题意:输入一个数字n,找出三个数字p1,p2,p3,满足p1<=p2<=p3并且p1+p2+p3=n
     1 //筛素数
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<algorithm>
     5 using namespace std;
     6 const int N = 10005;
     7 bool pri[N];
     8 int prime[N];
     9 int cnt;
    10 void init()
    11 {
    12     cnt = 0;
    13     pri[0] = pri[1] = 1;
    14     for(int i = 2; i < N; i++)
    15     {
    16         if(!pri[i]){
    17             prime[cnt++] = i;
    18             for(int j = i+i; j < N; j+=i)
    19             {
    20                 pri[j] = 1;
    21             }
    22         }
    23     }
    24     return;
    25 }
    26 int main()
    27 {
    28     int n;
    29     init();
    30     while(~scanf("%d",&n))
    31     {
    32         int ans = 0;
    33 
    34         //printf("%d
    ",cnt);
    35         for(int i = 0; i < cnt; i++)
    36         {
    37             if(3*prime[i]>n) break;
    38             for(int j = i; j < cnt; j++)
    39             {
    40                 if(prime[i]+2*prime[j]>n) break;
    41                 //for(int k = j; k < cnt; k++)
    42                 //{
    43                 //    if(prime[i]+prime[j]+prime[k]==n) ans++;// printf("%d %d %d
    ",prime[i],prime[j],prime[k]);printf("%d %d %d
    ",i,j,k);}
    44                // }
    45                if(!pri[n-prime[i]-prime[j]]) ans++;
    46             }
    47         }
    48         printf("%d
    ",ans);
    49     }
    50     return 0;
    51 }
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  • 原文地址:https://www.cnblogs.com/shanyr/p/5668988.html
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