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  • poj 2739 Sum of Consecutive Prime Numbers (尺取法)

    Sum of Consecutive Prime Numbers
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 21247   Accepted: 11625

    Description

    Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two representations 5 + 7 + 11 + 13 + 17 and 53. The integer 41 has three representations 2+3+5+7+11+13, 11+13+17, and 41. The integer 3 has only one representation, which is 3. The integer 20 has no such representations. Note that summands must be consecutive prime 
    numbers, so neither 7 + 13 nor 3 + 5 + 5 + 7 is a valid representation for the integer 20. 
    Your mission is to write a program that reports the number of representations for the given positive integer.

    Input

    The input is a sequence of positive integers each in a separate line. The integers are between 2 and 10 000, inclusive. The end of the input is indicated by a zero.

    Output

    The output should be composed of lines each corresponding to an input line except the last zero. An output line includes the number of representations for the input integer as the sum of one or more consecutive prime numbers. No other characters should be inserted in the output.

    Sample Input

    2
    3
    17
    41
    20
    666
    12
    53
    0

    Sample Output

    1
    1
    2
    3
    0
    0
    1
    2

    一开始迷之wa...
    先找出素数下标的上界就可以A...
    然后纠结了20分钟...
    然后发现是预处理的素数少了一个素数..
    我预处理是处理到<10005的素数...
    最大数10000,而超过10000的第一个素数是10007
    这样判断终止条件就会死循环...
    sad
     1 /*************************************************************************
     2     > File Name: code/poj/2739.cpp
     3     > Author: 111qqz
     4     > Email: rkz2013@126.com 
     5     > Created Time: 2015年09月25日 星期五 01时32分43秒
     6  ************************************************************************/
     7 
     8 #include<iostream>
     9 #include<iomanip>
    10 #include<cstdio>
    11 #include<algorithm>
    12 #include<cmath>
    13 #include<cstring>
    14 #include<string>
    15 #include<map>
    16 #include<set>
    17 #include<queue>
    18 #include<vector>
    19 #include<stack>
    20 #include<cctype>
    21 #define y1 hust111qqz
    22 #define yn hez111qqz
    23 #define j1 cute111qqz
    24 #define ms(a,x) memset(a,x,sizeof(a))
    25 #define lr dying111qqz
    26 using namespace std;
    27 #define For(i, n) for (int i=0;i<int(n);++i)  
    28 typedef long long LL;
    29 typedef double DB;
    30 const int inf = 0x3f3f3f3f;
    31 int pri[10005];
    32 int n ;
    33 int mx;
    34 bool prime ( int n)
    35 {
    36     if (n<=3) return true;
    37     for ( int i = 2 ; i*i<= n ; i++)
    38     {
    39     if (n%i==0) return false;
    40     }
    41     return true;
    42 }
    43 
    44 void solve()
    45 {
    46     int head = 1;
    47     int tail = 1;
    48     int sum = 0 ;
    49     int ans = 0 ;
    50     while (pri[tail]<=n)
    51     {
    52 //    cout<<"asd"<<endl;
    53     sum = sum + pri[tail];
    54     if (sum>=n)
    55     {
    56         while (sum>n)
    57         {
    58         sum = sum - pri[head];
    59         head++;
    60         }
    61     
    62         if (sum==n)
    63         {
    64         ans++;
    65         }
    66     }
    67        tail++;
    68     }
    69     printf("%d
    ",ans);
    70 }
    71 int main()
    72 {
    73   #ifndef  ONLINE_JUDGE 
    74   // freopen("in.txt","r",stdin);
    75   #endif
    76     
    77    int cnt = 0 ;
    78    for ( int i = 2 ; i <= 10011 ; i++)
    79     {
    80     if (prime(i))
    81     {
    82         cnt++;
    83         pri[cnt] = i ;
    84 //        if (i>10000) cout<<i<<endl;
    85     }
    86     }
    87     while (scanf("%d",&n)!=EOF&&n)
    88     {
    89     solve();
    90     }
    91    
    92  #ifndef ONLINE_JUDGE  
    93   fclose(stdin);
    94   #endif
    95     return 0;
    96 }
    View Code
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  • 原文地址:https://www.cnblogs.com/111qqz/p/4837940.html
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