Question

如何在一堆数字中找出几个数，这几个数的和能够相加得一个固定值

比如说，我有24个数，我要从里面找出4个，使得这4个数的值相加等于一个固定值（这个值我自己定）
如何编程？求高手！！！！！我把数字贴出来，能帮我做的就顺便啦，万分感谢！！！正确答案立马采纳！！
数值个数：25 要求找出4个，使得4个值相加等于5326.49
11.61

25.38

44.8

59.25

121.66

126.93

142.79

155.07

246.24

288.44

389.46

451.14

472.22

557.47

638.59

688.62

746.72

891.7

979.14

1028.66

1038.21

1153.38

1634.87

2760.68

Answer 1

In[6]:= a = {1161, 2538, 4480, 5925, 12166, 12693, 14279, 15507,
24624, 28844, 38946, 45114, 47222, 55747, 63859, 68862, 74672,
89170, 97914, 102866, 103821, 115338, 163487,
276068}; mb = 487235; aa =
Table[ToExpression[StringJoin["a", ToString[i]]], {i, 1, Length[a]}];
bb = a.aa; t =
Solve[Flatten[{Total[bb] == mb, Map[{# >= 0, # <= 1} &, aa]}], aa,
Integers]; Length[t]
aa /. t
Out[7]= 16
Out[8]= {{0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
1, 0, 0, 1}, {0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1,
1, 1, 0, 0, 0, 0}, {0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0,
0, 0, 1, 0, 0, 0, 1, 0}, {0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0,
1, 1, 0, 1, 0, 1, 1, 0, 0, 0}, {0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1,
0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0}, {1, 0, 0, 0, 1, 0, 0, 1, 1, 1,
0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0}, {1, 0, 0, 1, 1, 0, 0, 0,
0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0}, {1, 0, 0, 1, 1, 0,
0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0}, {1, 0, 1, 0,
0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0}, {1, 0,
1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1,
0}, {1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0,
0, 0, 0}, {1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0,
0, 1, 0, 0, 0}, {1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0,
1, 1, 0, 1, 0, 0, 0}, {1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0,
1, 0, 0, 0, 1, 1, 1, 0, 0}, {1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0,
0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0}, {1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1,
0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0}}
用Mathematica解出来有16组解，如果是4张，那就是第一个解。运算过程中我把数据全放大了100倍，用整数来运算。

Answer 2

遍历即可，我用matlab编的简单程序，其他语言类似

A=[11.61,25.38,44.8,59.25, 121.66,126.93 ,142.79,155.07,246.24,288.44,389.46,451.14,472.22,557.47,638.59,688.62,746.72,891.7,979.14,1028.66,1038.21,1153.38,1634.87,2760.68];
n=length(A); %求A中元素的个数
for i=1:n-3
for j=i+1:n-2
for k=j+1:n-1
for l=k+1:n
if A(i)+A(j)+A(k)+A(l)==4872.35 %判断条件
[A(i) A(j) A(k) A(l)]
end
end
end
end
end

我这里预设的和是4872.35
程序运行的结果是44.8、1028.66、1038.21、2760.68
你要的和无解

Answer 3

C++版代码如下，做了一点拓展，可以求出通解
#include<iostream>
#include<cstdio>
using namespace std;
double a[1000], b[1000], c[1000];
double x ;
int n,m ;

void f(int i, int l){
if(i > n || l > m) return;
b[l] = a[i];
c[l] = i ;

if(l == m){
double sum = 0;
for(int k = 1; k <= m; k++) sum += b[k];

if(sum == x){
for(int k = 1 ; k <= m ; k++){
cout << b[k];
if(k < m) cout << " + ";
}
cout << " = " << x << endl << "这是";
for(int k = 1 ; k <= m ; k++){
cout << "第 " << c[k] << " 个数" ;
if(k < m) cout << "、";
}
cout << "相加。" << endl;
}
}

else{ for(int y = i + 1; y <= n - m + l + 1 ; y ++) f(y, l+1); }

return;
}

int main(){
cout << "请输入数字个数（完成后回车）：";
cin >> n;

cout << "请输入这"<<n<<"个数，每输入一个间隔一个空格（所有数字都输入完成后回车）：\n" ;
for(int i = 1; i <= n ; i++) cin >> a[i];

cout <<"请输入需要选出多少个数相加（完成后回车）\n";
cout <<"（若需要的是选出任意个数的，请输入0然后回车）：\n";
cin >> m ;

cout << "请输入所需得到的固定值（完成后回车）：" ;
cin >> x ;

if(m == 0){
for(m = 1; m <= n; m++){
cout << "\n选出" << m << "个数相加，所得到的组合有：\n";
for(int i = 1; i <= n-m+1; i++) f(i,1);
}
}

else{
cout << "\n选出" << m << "个数相加，所得到的组合有：\n";
for(int i = 1; i <= n-m+1; i++) f(i,1);
}

return 0;
}

Answer 4

你要的和不存在，。java版如下:

public class test {

public static void main(String[] args) {
Double[] d={11.61,25.38,44.8,59.25,121.66,126.93,142.79,155.07,246.24,288.44,389.46,451.14,472.22,557.47,638.59,688.62,746.72 , 891.7,979.14,1028.66,1038.21,1153.38,1634.87,2760.68};

double s=0.0;
int count=0;
double a=5560.54;
for(int i=0;i<24;i++) {
for(int j=1;j<24;j++) {
for(int x=2;x<24;x++) {
for(int y=3;y<24;y++) {
s=d[i]+d[j]+d[x]+d[y];

if(s==a){

System.out.print(d[i]+",");
System.out.print(d[j]+",");
System.out.print(d[x]+",");
System.out.println(d[y]);
System.out.println("------");

}
count=count+1;

}
}
}
}
System.out.println(count);
}

}