poj2739解题报告
2014-10-31 23:52
363 查看
Sum of Consecutive Prime Numbers
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
Sample Output
源代码
#include<iostream>
#include<math.h>
#include<string>
using namespace std;
int main()
{
int i,j,n,t=0,temp,sum,num,s;
int a[2000];
for (i=2;i<=10000;i++)
{
temp=0;
if (i%2==0 && i!=2) {continue;}
for (j=2;j<i;j++)
{
if (i%j==0) {temp=1;}
}
if (temp==0 || i==2) {a[t]=i;t++;}
}
for (;;)
{
cin>>n;
num=0;
for (i=0;i<t;i++)
{
if (n>=a[i] && n<=a[i+1]) {s=i+1;}
}
if (n==0) return 0;
else
{
for(i=0;i<s;i++)
{
sum=0;
if(a[i]>n) break;
for(j=i;j<t;j++)
{
sum+=a[j];
if(sum==n)
{
num++;
break;
}
if (sum>n)
break;
}
}
cout<<num<<endl;
}
}
return 0;
}
本人一开始把所有素数列出再枚举,试了好几次- -,都是超时,用了动态规划,把其中肯定不是素数的直接排除掉,我就排除了2的倍数,速度马上飞了上去,本题不算太难,细心即可a掉!
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 19540 | Accepted: 10720 |
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
源代码
#include<iostream>
#include<math.h>
#include<string>
using namespace std;
int main()
{
int i,j,n,t=0,temp,sum,num,s;
int a[2000];
for (i=2;i<=10000;i++)
{
temp=0;
if (i%2==0 && i!=2) {continue;}
for (j=2;j<i;j++)
{
if (i%j==0) {temp=1;}
}
if (temp==0 || i==2) {a[t]=i;t++;}
}
for (;;)
{
cin>>n;
num=0;
for (i=0;i<t;i++)
{
if (n>=a[i] && n<=a[i+1]) {s=i+1;}
}
if (n==0) return 0;
else
{
for(i=0;i<s;i++)
{
sum=0;
if(a[i]>n) break;
for(j=i;j<t;j++)
{
sum+=a[j];
if(sum==n)
{
num++;
break;
}
if (sum>n)
break;
}
}
cout<<num<<endl;
}
}
return 0;
}
本人一开始把所有素数列出再枚举,试了好几次- -,都是超时,用了动态规划,把其中肯定不是素数的直接排除掉,我就排除了2的倍数,速度马上飞了上去,本题不算太难,细心即可a掉!
相关文章推荐
- POJ2739解题报告
- poj 2739解题报告
- poj2739 解题报告(poj 2739 analysis report)
- 【原】 POJ 2739 Sum of Consecutive Prime Numbers 筛素数+积累数组 解题报告
- POJ 2739 解题报告
- POJ 2739(连续素数和) 解题报告
- poj 2739 Sum of Consecutive Prime Numbers 解题报告
- POJ 2739(连续素数和) 解题报告
- LA-3399 & POJ-2739 Sum of Consecutive Prime Numbers 解题报告
- poj2498解题报告
- POJ 1797 Heavy Transportation(Dijkstra变形) 解题报告
- poj解题报告——1003、1005、1799
- POJ 2942 Tarjan双联通分量+二分图 解题报告
- poj 2449 Remmarguts' Date 第k短路 A*+spfa 解题报告
- POJ-1411 & HDOJ-1239 Calling Extraterrestrial Intelligence Again 解题报告
- POJ - 1442 Black Box解题报告(求第k小的数 堆)
- POJ1068 解题报告
- POJ 2719 Faulty Odometer 解题报告
- POJ - 1195 Mobile phones解题报告(二维树状数组)
- POJ 3278解题报告(C语言版)//Catch That Cow