Prime Bases
题目连接：
Descriptionww.co
Given any integer base b >= 2, it is well known that every positive integer n can be uniquely represented in base b. That is, we can write
n = a0 + a1* b + a2* b* b + a3* b* b* b + ...
where the coefficients a0, a1, a2, a3, ... are between 0 and b-1 (inclusive).
What is less well known is that if p0, p1, p2, ... are the first primes (starting from 2, 3, 5, ...), every positive integer n can be represented uniquely in the "mixed" bases as:
n = a0 + a1* p0 + a2* p0* p1 + a3* p0* p1* p2 + ...
where each coefficient ai is between 0 and pi-1 (inclusive). Notice that, for example, a3 is between 0 and p3-1, even though p3 may not be needed explicitly to represent the integer n.
Given a positive integer n, you are asked to write n in the representation above. Do not use more primes than it is needed to represent n, and omit all terms in which the coefficient is 0.
Input
Each line of input consists of a single positive 32-bit signed integer. The end of input is indicated by a line containing the integer 0.
Output
For each integer, print the integer, followed by a space, an equal sign, and a space, followed by the mixed base representation of the integer in the format shown below. The terms should be separated by a space, a plus sign, and a space. The output for each integer should appear on its own line.
Sample Input
123
456
123456
0
Sample Output
123 = 1 + 12 + 4235
456 = 123 + 1235 + 22357
123456 = 123 + 6235 + 42357 + 1235711 + 4235711*13
Hint
题意
给你一个数，你需要拆成素数因子的形式
比如123 = 1 + 1*2+4*2*3*5
拆成n = a0 + a1* p0 + a2* p0* p1 + a3* p0* p1* p2 + ...
的形式
给你一个数，问你怎么拆
题解：
贪心去拆就好了让，素数乘积从大到小不断考虑
如果超过就减去就好了
然后不断贪
代码
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 30;
bool flag[MAXN];
vector<int> P;
void GetPrime_1()
{
int i, j;
memset(flag, false, sizeof(flag));
for (i = 2; i < MAXN; i++)
if (!flag[i])
{
P.push_back(i);
for (j = i; j < MAXN; j += i)
flag[j] = true;
}
}
string get(int p)
{
string k;
while(p)
{
k+=(char)(p%10+'0');
p/=10;
}
reverse(k.begin(),k.end());
return k;
}
int main()
{
GetPrime_1();
long long n;
while(cin>>n)
{
if(n==0)break;
long long ans = 1;
for(int i=0;i<P.size();i++)
ans*=P[i];
long long pre = n;
stack<int> S;
for(int i=P.size()-1;i>=0;i--)
{
S.push(n/ans);
n=n%ans;
ans/=P[i];
}
printf("%lld =",pre);
int first = 1;
if(n)
{
printf(" 1");
first = 0;
}
for(int i=0;i<P.size();i++)
{
if(S.top()==0)
{
S.pop();
continue;
}
if(!first)
printf(" +");
first=0;
printf(" %d",S.top());
S.pop();
for(int j=0;j<=i;j++)
{
printf("*%d",P[j]);
}
}
printf("\n");
}
}