zoukankan      html  css  js  c++  java
  • poj 2739 Sum of Consecutive Prime Numbers 素数 读题 难度:0

    Sum of Consecutive Prime Numbers
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 19697   Accepted: 10800

    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

    思路:打表,不管是dp还是简单的列举首尾端点
    问题:看错题意以为3 5 5 7 是允许的方式

    #include <cstdio>
    #include <cstring>
    using namespace std;
    int method[10001][1300];
    int dp[10001];
    bool isntprime[10001];
    int heap[1300],cnt;
    void calprime(){
        method[2][0]=1;
        dp[2]++;
        heap[cnt++]=2;
        for(int i=3;i<10001;i+=2){
            if(!isntprime[i]){
                heap[cnt]=i;
                method[i][cnt++]=1;dp[i]++;
                for(int j=3;i*j<10001;j+=2){
                    isntprime[i*j]=true;
                }
            }
        }
    }
    void caldp(){
        for(int i=2;i<10001;i++){
            for(int j=0;j<cnt;j++){
                if(method[i][j]!=0){
                    if(j<cnt-1&&i+heap[j+1]<10001){
                        method[i+heap[j+1]][j+1]+=method[i][j];
                        dp[i+heap[j+1]]+=method[i][j];
                    }
                }
            }
        }
    }
    int main(){
        calprime();
        caldp();
        int n;
        while(scanf("%d",&n)==1&&n){
            printf("%d
    ",dp[n]);
        }
        return 0;
    }
    

      

  • 相关阅读:
    SQL Server 中的事务与事务隔离级别以及如何理解脏读, 未提交读,不可重复读和幻读产生的过程和原因
    微软BI 之SSIS 系列
    微软BI 之SSIS 系列
    微软BI 之SSIS 系列
    微软BI 之SSIS 系列
    微软BI 之SSIS 系列
    微软BI 之SSAS 系列
    微软BI 之SSRS 系列
    微软BI 之SSRS 系列
    配置 SQL Server Email 发送以及 Job 的 Notification通知功能
  • 原文地址:https://www.cnblogs.com/xuesu/p/4115976.html
Copyright © 2011-2022 走看看