"codevs2549/2548" natural number and/or integral solution

Natural numbers and decomposition description

The natural number n is decomposed into the sum of several natural numbers, and the output scheme number.

Input

N, (1≤N≤50)

Output

Number of programmes

Sample Input

5

Sample Output

7

HINT

5 can be divided into

1 1 1) 1 1
1 1 1 2
1 1 3
1 2 2
1 4
2 3
5

Exercises

Solution 1: Full backpack

#include <iostream>#include<cstdio>using namespacestd;intN,ans;intA -]; intMain () {scanf ("%d",&N); a[0] =1;  for(intI=1; i<=n;i++)         for(intj=1; j<=n;j++)        {            if(j>=i) a[j]+ = a[j-i]; } printf ("%d", A[n]);}

Solution 2: Search

x indicates which number is subtracted from the previous step, and rest represents the remaining number.

Get a solution when rest is 0

I:x->rest promise not to repeat.

#include <iostream>#include<cstdio>using namespacestd;intAns,n;voidDfsintXintRS) {    if(rs = =0) {ans++;return; }     for(inti=x;i<=rs;i++) DFS (I,rs-i);}intMain () {scanf ("%d",&N); DFS (1, N); cout<<ans;}

Description of natural number integral solution

The natural number n is decomposed into the product of several natural numbers, and the output scheme number.

Input

Natural number n, (1≤n≤2000000000)

Output

Number of programmes

Sample Input

20

Sample Output

4

HINT

20 can be divided into

20
4 5
2 10
2 2 5

Exercises

As with the above question, when rest equals 1 o'clock, you get an answer.

60 min

#include <iostream>#include<cstdio>using namespacestd;intN,ans;voidDfsintXintRS) {    if(rs = =1) ans++;  for(inti=x;i<=rs;i++)        if(rs% i = =0&& i!=1) DFS (I,rs/i);}intMain () {scanf ("%d",&N); DFS (1, N); printf ("%d", ans);}

The problem data range is too large to be optimized.

When an i is found to make the RS die I=0, a set of solutions are found, and the approximations may continue to decompose

#include <iostream>#include<cstdio>#include<cmath>using namespacestd;intn,ans=1;voidDfsintXintRS) {     for(intI=x;i<=sqrt (RS); i++)        if(rs% i = =0) ans++,dfs (i,rs/i);}intMain () {scanf ("%d",&N); DFS (2, N); printf ("%d", ans);}

"codevs2549/2548" natural number and/or integral solution