找零錢所需最少張數
阿新 • • 發佈:2018-12-15
#include<iostream> #include<vector> #define INF 10000000; using namespace std; int changes(int a[],int len,int value) { int **dp=new int*[len+1]; for(int i=0;i<=len;++i) dp[i]=new int[value+1]; for(int i=1;i<=len;++i) dp[i][0]=0; for(int j=1;j<=value;++j) dp[0][j]=INF; for(int i=1;i<=len;++i) for(int j=1;j<=value;++j) { if(a[i]>j) dp[i][j]=dp[i-1][j]; else { dp[i][j] = dp[i][j-a[i]]+1 < dp[i-1][j] ? dp[i][j-a[i]]+1 : dp[i-1][j]; } } int result=dp[len][value]; for(int i=0;i<=len;++i) delete dp[i]; delete[] dp; return result; } int Changes(int change[],int len,int value) { int *dp=new int[value+1](); for(int i=0;i<value+1;++i) dp[i]=INF; dp[0]=0; for(int i=1;i<=len;++i) for(int j=change[i];j<=value;++j) { // dp[0]=1; // if(j>change[i]) dp[j]=dp[j-change[i]]+1 < dp[j] ? dp[j-change[i]]+1 : dp[j]; } int result=dp[value]; delete[] dp; return result; } int main() { int a[]={0,1,5,10,25}; int len=sizeof(a)/sizeof(a[0]); cout<<Changes(a,len,17)<<endl; }