邮票组合问题
有四种面值的邮票很多枚,面值分别为1,4,12,21,取五张,求取出这些邮票的最大连续组合值
代码:
#include <stdio.h>
#include <string.h>
#define N 5
#define M 5
int k, Found, Flag[N];
int Stamp[M] = {, , , , };
// 在剩余张数n中组合出面值和Value
int Combine(int n, int Value)
{
if(n >= && Value == )
{
Found = ;
int Sum = ;
for(int i=; i<N && Flag[i] != ; i++)
{
Sum += Stamp[Flag[i]];
printf("%d ", Stamp[Flag[i]]);
}
printf("\tSum=%d\n\n", Sum);
}
else
{
for(int i=; i<M && !Found && n>; i++)
if(Value-Stamp[i] >= )
{
Flag[k++] = i;
Combine(n-, Value-Stamp[i]);
Flag[--k] = ;
}
}
return Found;
}
int main(int argc, char* argv[])
{
for(int i=; Combine(N, i); i++, Found=);
}
改进算法:
#include <iostream>
#include <string.h>
using namespace std;#define MAX 10
#define MAXINT 100000#define min(a, b) ((a) >= (b) ? (b) : (a))int Stamp[MAX+];
int StampCnt[MAXINT] = {};int MaxValue(int n, int m)
{
int i, j; for(i = ; ; i++)
{
StampCnt[i] = ;
for(j = ; j <= m; j++)
{
if(Stamp[j] == i)
{
StampCnt[i] = ;
break;
}
else if(i > Stamp[j])
{
StampCnt[i] = min(StampCnt[i], StampCnt[i-Stamp[j]] + );
}
} if(StampCnt[i] > n)
return i-;
}
// return 0;
}int main()
{
int N, M;
int i, j; cin >> N >> M; //5 5, 输入 1 2 4 12 21, 输出 71 memset(Stamp, , sizeof(Stamp));
for(i = ; i <= M; i++)
{
cin >> Stamp[i];
} cout << MaxValue(N, M) << endl;
}