【簡單遞歸】
題目描述
已知 n 個整數b1,b2,…,bn
以及一個整數 k（k＜n）。
從 n 個整數中任選 k 個整數相加，可分別得到一系列的和。
例如當 n=4，k＝3，4 個整數分別為 3，7，12，19 時，可得全部的組合與它們的和為：
3＋7＋12=22 3＋7＋19＝29 7＋12＋19＝38 3＋12＋19＝34。
現在，要求你計算出和為素數共有多少種。
例如上例，只有一種的和為素數：3＋7＋19＝29。
輸入
第一行兩個整數：n , k （1<=n<=20，k＜n）
第二行n個整數：x1,x2，…,xn （1<=xi<=5000000）
輸出
一個整數（滿足條件的方案數）。
樣例輸入
4 3
3 7 12 19
樣例輸出
1
代碼
#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
//判斷是否是素數
bool isprime(int n)
{
if (n<1)return false;
for (int i=2;i*i<n;i++)
{
if (n%i==0)//n&(i-1)==0
return false;
}
return true;
}
int a[22],p[22], b[22];
bool vis[22];
int n, k, sum, ans;
int cnt = 0;
vector<vector<int> > index_vec;
vector<int> sum_vec;
void dfs(int index)
{
cout << "Enter dfs, " << ++cnt << "th" << ", index:" << index << ", p[index-1]:" << p[index-1] << endl;
if (index==k+1)
{
if (isprime(sum))
{
ans++;
cout << "sum:" << sum << " is prime." << endl;
}
//打印每一個元素組合
vector<int> v;
for (int i=1;i<=index-1;i++)
{
cout << p[i] << " ";
v.push_back(p[i]);
}
index_vec.push_back(v);
sum_vec.push_back(sum);
cout << endl;
cout << "sum:" << sum << endl;
return;
}
//遍歷n個數
for (int i=1;i<=n;i++)
{
//條件1：該元素在本次遍歷中是否被訪問過
//條件2：保障遍歷的i下標大於已統計過的最后一個元素下標
cout << "i:" << i << ", vis:" << vis[i] << ", index:" << index << ", p[index-1]:" << p[index-1] << endl;
if (vis[i]==false && i>p[index-1])
{
p[index] = i;//數組p用來保存元素下標，index為當前統計的元素總數目
vis[i] = true;//標記該元素已被該次計算統計過
sum += a[i];//更新和,在全排列過程中，對結果進行統計
cout << "Enter the next dfs" << endl;
cout << "sum:" << sum << ", i:" << i << ", vis[i]:" << vis[i] << endl;
dfs(index+1);//開始下一次遞歸計算
cout << "return from dfs" << endl;
//回溯
vis[i] = false;
sum -= a[i];
cout << "sum:" << sum << ", i:" << i << ", vis[i]:" << vis[i] << ", index:" << index << ", p[index-1]:" << p[index-1] << endl;
}
}
}
int main()
{
//初始化全局變量
memset(b, 0, sizeof(b));
memset(vis, false, sizeof(vis));
//獲取輸入
cin >> n >> k;
cout << "n=" << n << " ,k=" << k << endl;
for (int i=1; i<=n;i++)
{
cin >> a[i];
p[i]=i;
}
ans = 0;
dfs(1);
cout << ans << endl;
int seq=0;
for (auto it: sum_vec)
{
cout << "sum: " << it << endl;
for (auto it_idx: index_vec[seq])
{
cout << it_idx << " ";
}
cout << endl;
++seq;
}
return 0;
}
題目解析
循環中調用遞歸
代碼打印幫助理解遞歸過程
4
3
n=4 ,k=3
3
7
12
19
Enter dfs, 1th, index:1, p[index-1]:0
i:1, vis:0, index:1, p[index-1]:0
Enter the next dfs
sum:3, i:1, vis[i]:1
Enter dfs, 2th, index:2, p[index-1]:1
i:1, vis:1, index:2, p[index-1]:1
i:2, vis:0, index:2, p[index-1]:1
Enter the next dfs
sum:10, i:2, vis[i]:1
Enter dfs, 3th, index:3, p[index-1]:2
i:1, vis:1, index:3, p[index-1]:2
i:2, vis:1, index:3, p[index-1]:2
i:3, vis:0, index:3, p[index-1]:2
Enter the next dfs
sum:22, i:3, vis[i]:1
Enter dfs, 4th, index:4, p[index-1]:3
1 2 3
sum:22
return from dfs
sum:10, i:3, vis[i]:0, index:3, p[index-1]:2
i:4, vis:0, index:3, p[index-1]:2
Enter the next dfs
sum:29, i:4, vis[i]:1
Enter dfs, 5th, index:4, p[index-1]:4
sum:29 is prime.
1 2 4
sum:29
return from dfs
sum:10, i:4, vis[i]:0, index:3, p[index-1]:2
return from dfs
sum:3, i:2, vis[i]:0, index:2, p[index-1]:1
i:3, vis:0, index:2, p[index-1]:1
Enter the next dfs
sum:15, i:3, vis[i]:1
Enter dfs, 6th, index:3, p[index-1]:3
i:1, vis:1, index:3, p[index-1]:3
i:2, vis:0, index:3, p[index-1]:3
i:3, vis:1, index:3, p[index-1]:3
i:4, vis:0, index:3, p[index-1]:3
Enter the next dfs
sum:34, i:4, vis[i]:1
Enter dfs, 7th, index:4, p[index-1]:4
1 3 4
sum:34
return from dfs
sum:15, i:4, vis[i]:0, index:3, p[index-1]:3
return from dfs
sum:3, i:3, vis[i]:0, index:2, p[index-1]:1
i:4, vis:0, index:2, p[index-1]:1
Enter the next dfs
sum:22, i:4, vis[i]:1
Enter dfs, 8th, index:3, p[index-1]:4
i:1, vis:1, index:3, p[index-1]:4
i:2, vis:0, index:3, p[index-1]:4
i:3, vis:0, index:3, p[index-1]:4
i:4, vis:1, index:3, p[index-1]:4
return from dfs
sum:3, i:4, vis[i]:0, index:2, p[index-1]:1
return from dfs
sum:0, i:1, vis[i]:0, index:1, p[index-1]:0
i:2, vis:0, index:1, p[index-1]:0
Enter the next dfs
sum:7, i:2, vis[i]:1
Enter dfs, 9th, index:2, p[index-1]:2
i:1, vis:0, index:2, p[index-1]:2
i:2, vis:1, index:2, p[index-1]:2
i:3, vis:0, index:2, p[index-1]:2
Enter the next dfs
sum:19, i:3, vis[i]:1
Enter dfs, 10th, index:3, p[index-1]:3
i:1, vis:0, index:3, p[index-1]:3
i:2, vis:1, index:3, p[index-1]:3
i:3, vis:1, index:3, p[index-1]:3
i:4, vis:0, index:3, p[index-1]:3
Enter the next dfs
sum:38, i:4, vis[i]:1
Enter dfs, 11th, index:4, p[index-1]:4
2 3 4
sum:38
return from dfs
sum:19, i:4, vis[i]:0, index:3, p[index-1]:3
return from dfs
sum:7, i:3, vis[i]:0, index:2, p[index-1]:2
i:4, vis:0, index:2, p[index-1]:2
Enter the next dfs
sum:26, i:4, vis[i]:1
Enter dfs, 12th, index:3, p[index-1]:4
i:1, vis:0, index:3, p[index-1]:4
i:2, vis:1, index:3, p[index-1]:4
i:3, vis:0, index:3, p[index-1]:4
i:4, vis:1, index:3, p[index-1]:4
return from dfs
sum:7, i:4, vis[i]:0, index:2, p[index-1]:2
return from dfs
sum:0, i:2, vis[i]:0, index:1, p[index-1]:0
i:3, vis:0, index:1, p[index-1]:0
Enter the next dfs
sum:12, i:3, vis[i]:1
Enter dfs, 13th, index:2, p[index-1]:3
i:1, vis:0, index:2, p[index-1]:3
i:2, vis:0, index:2, p[index-1]:3
i:3, vis:1, index:2, p[index-1]:3
i:4, vis:0, index:2, p[index-1]:3
Enter the next dfs
sum:31, i:4, vis[i]:1
Enter dfs, 14th, index:3, p[index-1]:4
i:1, vis:0, index:3, p[index-1]:4
i:2, vis:0, index:3, p[index-1]:4
i:3, vis:1, index:3, p[index-1]:4
i:4, vis:1, index:3, p[index-1]:4
return from dfs
sum:12, i:4, vis[i]:0, index:2, p[index-1]:3
return from dfs
sum:0, i:3, vis[i]:0, index:1, p[index-1]:0
i:4, vis:0, index:1, p[index-1]:0
Enter the next dfs
sum:19, i:4, vis[i]:1
Enter dfs, 15th, index:2, p[index-1]:4
i:1, vis:0, index:2, p[index-1]:4
i:2, vis:0, index:2, p[index-1]:4
i:3, vis:0, index:2, p[index-1]:4
i:4, vis:1, index:2, p[index-1]:4
return from dfs
sum:0, i:4, vis[i]:0, index:1, p[index-1]:0
1
sum: 22
1 2 3
sum: 29
1 2 4
sum: 34
1 3 4
sum: 38
2 3 4
Process finished with exit code 0