如何解决将特定数字相加的不同方法获得100
| 我想编写一个代码，说明一个人可以通过多种方式将5个不同的数字求和得到100。例如，数字为2,5,10,20,50
50+50
20+20+20+20+20
#include<iostream>
#include<vector>
using namespace std;
int i,sum,n=5,counter=0;
int add(vector<int> &m){
if(m.size()==0) return 0 ;
for(i=0 ; i<m.size() ; i++ ){
sum=m[i]+add(m);
cout<< sum<<endl;
if(n>0) n--;
m.resize(n);
}
}
int _tmain(int argc,_TCHAR* argv[])
{
int i,n=5;
vector<int> m;
m.resize(5);
m[0]=2;
m[1]=5;
m[2]=10;
m[3]=20;
m[4]=50;
add(m);
return 0;
}
解决方法
只是为了好玩
#include <iostream>
#include <vector>
#include <iterator>
#include <numeric>
#include <algorithm>
static const int terms[] = { 2,5,10,20,50,/*end marker*/0 };
using namespace std;
typedef vector <int> Solution;
typedef vector <Solution> Solutions;
inline int Sum(const Solution& s)
{
return accumulate(s.begin(),s.end(),0);
}
template <typename OutIt>
OutIt generate(const int target,const int* term,Solution partial,OutIt out)
{
const int cumulative = Sum(partial); // TODO optimize
if (cumulative>target)
return out; // bail out,target exceeded
if (cumulative == target)
{
(*out++) = partial; // report found solution
return out;
} else
{
// target not reached yet,try all terms in succession
for (; *term && cumulative+*term<=target; term++)
{
partial.push_back(*term);
out = generate(target,term,partial,out); // recursively generate till target reached
partial.pop_back();
}
return out;
}
}
Solutions generate(const int target)
{
Solutions s;
generate(target,terms,Solution(),back_inserter(s));
return s;
}
void Dump(const Solution& solution)
{
std::copy(solution.begin(),solution.end(),std::ostream_iterator<int>(std::cout,\" \"));
std::cout << std::endl;
}
#ifdef _TCHAR
int _tmain(int argc,_TCHAR* argv[])
#else
int main(int argc,char* argv[])
#endif
{
Solutions all = generate(100);
for_each(all.rbegin(),all.rend(),&Dump);
return 0;
}
target=2000
#include <iostream>
#include <vector>
using namespace std;
size_t generate(const int target,vector<int> terms)
{
size_t count = 0;
if (terms.back()<=target)
{
int largest = terms.back();
terms.pop_back();
int remain = target % largest;
if (!remain)
count += 1;
if (!terms.empty())
for (; remain<=target; remain+=largest)
count += generate(remain,terms);
}
return count;
}
int main(int argc,char* argv[])
{
static const int terms[] = {2,50};
std::cout << \"Found: \" << generate(1000,vector<int>(terms,terms+5)) << std::endl;
return 0;
}
x^100
1/(1-x^2) * 1/(1-x^5) * 1/(1-x^10) * 1/(1-x^20) * 1/(1-x^50)
f(x) = 1/(1-x^A) * 1/(1-x^B) * ... * 1/(1-x^N)
f(x)
x^M
M
M = a*A + b*B + ... + n*N
a,b,...,n
f(x)
1/(1-x^A)
a
x^(a*A)
#include <iostream>
template<size_t N>
void find_combinations_helper(int total,const int (&denoms)[N],int denoms_used,int remaining,int (&counts)[N])
{
if (remaining == 0) {
int partial_sum = 0;
for( int i = 0; i < denoms_used; ++i ) {
if (counts[i]) {
std::cout << counts[i] << \"*\" << denoms[i];
partial_sum += counts[i] * denoms[i];
if (partial_sum < total) std::cout << \" + \";
}
}
std::cout << \"\\n\";
return;
}
if (denoms_used == N) return;
for( counts[denoms_used] = 0; remaining >= 0; (remaining -= denoms[denoms_used]),++counts[denoms_used] )
find_combinations_helper(total,denoms,denoms_used + 1,remaining,counts);
}
template<size_t N>
void find_combinations( int total,const int (&denoms)[N] )
{
int solutions[N];
find_combinations_helper(total,total,solutions);
}
int main(void) {
const int bill_denoms[] = { 50,2 };
find_combinations(100,bill_denoms);
}
m[0]=2;
...
m[0]=50;
add
m
add(m)
add
#include <iostream>
#include <sstream>
#include <vector>
void add(int i,std::string s,int sum)
{
if (sum == 100)
{
std::cout << s << \"=100\" << std::endl;
return;
}
if (sum > 100)
{
return;
}
if (sum < 100)
{
std::ostringstream oss;
oss << s << \"+\" << i;
add(i,oss.str(),sum+i);
}
}
int main()
{
std::vector<int> m;
m.resize(5);
m[0]=2;
m[1]=5;
m[2]=10;
m[3]=20;
m[4]=50;
// This loop will initiate m.size lines of recursive calls
// one for each element of the array
for (size_t i = 0; i < m.size(); i++)
{
add(m[i],\"\",0);
}
return 0;
}
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