POJ 1365 Prime Land
Prime Land
Time Limit: 1000MS | Memory Limit: 10000K | |
|---|---|---|
Total Submissions: 2122 | Accepted: 979 |
Description
Everybody in the Prime Land is using a prime base number system. In this system, each positive integer x is represented as follows: Let {pi}i=0,1,2,... denote the increasing sequence of all prime numbers. We know that x > 1 can be represented in only one way in the form of product of powers of prime factors. This implies that there is an integer kx and uniquely determined integers ekx, ekx-1, ..., e1, e0, (ekx > 0), that
The sequence (ekx, ekx-1, ... ,e1, e0) is considered to be the representation of x in prime base number system. It is really true that all numerical calculations in prime base number system can seem to us a little bit unusual, or even hard. In fact, the children in Prime Land learn to add to subtract numbers several years. On the other hand, multiplication and division is very simple. Recently, somebody has returned from a holiday in the Computer Land where small smart things called computers have been used. It has turned out that they could be used to make addition and subtraction in prime base number system much easier. It has been decided to make an experiment and let a computer to do the operation ``minus one''. Help people in the Prime Land and write a corresponding program. For practical reasons we will write here the prime base representation as a sequence of such pi and ei from the prime base representation above for which ei > 0. We will keep decreasing order with regard to pi.
Input
The input consists of lines (at least one) each of which except the last contains prime base representation of just one positive integer greater than 2 and less or equal 32767. All numbers in the line are separated by one space. The last line contains number 0.
Output
The output contains one line for each but the last line of the input. If x is a positive integer contained in a line of the input, the line in the output will contain x - 1 in prime base representation. All numbers in the line are separated by one space. There is no line in the output corresponding to the last ``null'' line of the input.
Sample Input
17 1
5 1 2 1
509 1 59 1
0Sample Output
2 4
3 2
13 1 11 1 7 1 5 1 3 1 2 1题意：sum=(x1^n1)*(x2^n2)......(xn^nn) 求sum-1的因式分解
通过这道题学会了sscanf将字符窜转换成数字
sscanf(str,"%d",&t);//str是字符窜，t要赋予地址
还有个strtok，这个函数不错以后可以用的着
http://www.cppblog.com/masiyou/archive/2009/10/07/98038.html#include <string.h>
#include <stdio.h>
int main(void)
{
char input[16] = "abc,dhh,eee";
char *p;
/* strtok places a NULL terminator
in front of the token, if found */
p = strtok(input, ",");
if (p) printf("%s\n", p);
/* A second call to strtok using a NULL
as the first parameter returns a pointer
to the character following the token */
p = strtok(NULL, ",");
if (p) printf("%s\n", p);
p = strtok(NULL, ",");
if (p) printf("%s\n", p);
return 0;
}#include <stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int MAXN=40000;
struct Node
{
int a,b;
}rem[1000];
char ss[1000];
int prime[MAXN],num[1000],vis[MAXN];
char str[1000];
int cnt,tes;
void init()
{
int i,j;
tes=0;
memset(vis,0,sizeof(vis));
for(i=2; i<MAXN; i++)
{
if(vis[i]==0)
{
for(j=i+i; j<MAXN; j+=i)
{
vis[j]=1;
}
}
}
tes=0;
for(i=2; i<MAXN; i++)
{
if(vis[i]==0) prime[tes++]=i;
}
}
void get_num()
{
int tot,i;
cnt=tot=0;
int t;
for(i=0; ss[i]; i++)
{
if(ss[i]==' ')
{
str[tot]='\0';
sscanf(str,"%d",&t);
num[cnt++]=t;
tot=0;
}
else str[tot++]=ss[i];
}
}
bool cmp(Node a,Node b)
{
return a.a>b.a;
}
int main()
{
int i,ans,cas,flag,cot;
init();
while(gets(ss))
{
flag=0;
if(ss[0]=='0') break;
int len=strlen(ss);
ss[len]=' ';
ss[len+1]='\0';
get_num();
ans=1;
for(i=0; i<cnt; i+=2)
{
ans*=pow(num[i],num[i+1]);
}
ans-=1;
cot=0;
for(i=0; i<tes; i++)
{
if(ans==1) break;
if(ans%prime[i]==0)
{
rem[cot].a=prime[i];
}
cas=0;
while(ans%prime[i]==0)
{
cas++;
ans/=prime[i];
}
if(cas) rem[cot++].b=cas;
}
sort(rem,rem+cot,cmp);
for(i=0;i<cot;i++)
{
if(i==0)printf("%d %d",rem[i].a,rem[i].b);
else printf(" %d %d",rem[i].a,rem[i].b);
}
printf("\n");
}
return 0;
}腾讯云开发者
