Nyoj 571--Various Division numbers

Integer division (c) time limit:MS | Memory limit:65535 KB  

Describe

Integer partitioning is a classic problem. Please write a program to complete the following requirements.

Each set of inputs is two integers n and K. (1 <= n <=, 1 <= k <= N)

output

for the input n,k;
The first line: divides n into the number of divisions of a number of positive integers.
Second line: divides n into the sum of k positive integers.
Line three: divides n into the number of divisions that do not exceed the maximum number of K.
Line Fourth: divides n into a number of the sum of the odd positive integers.
Line Fifth: divides n into the sum of a number of different integers.
Line Sixth: Print a blank line

 5 2 

 72333 

prompt

sample output hint: Br>1. Divide 5 into the sum of several positive integers as: 5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1
2. Divides 5 into 2 positive integers: 3+2, 4+1
3. Divide 5 into Maximum number Over 2 is divided into: 1+1+1+1+1, 1+1+1+2, 1+2+2
4. Divides 5 into the sum of several odd positive integers into: 5, 1+1+3, 1+1+1+1+1
5. Divides 5 into a number of different integers: 5, 1+4, 2+3

#include <cstdio>#include<iostream>#include<cstring>#include<string>using namespacestd;intdp[ -][ -];intMain () {intN, K, A, B, C, D, E;  while(SCANF ("%d%d", &n, &k)! =EOF) {        //Dividing the positive integer I into the number of positive integers not exceeding J         for(intI=1; i<=n; i++) {             for(intj=1; j<=n; J + +) {                if(j>i) Dp[i][j] =Dp[i][i]; Else if(j = i) dp[i][j] = dp[i][j-1] +1; ElseDP[I][J] = dp[i][j-1] + dp[i-J]            [j]; }        }A =Dp[n][n]; C=Dp[n][k]; //Dividing a positive integer into a number of k positive integers         for(intI=1; i<=n; i++) {             for(intj=1; j<=k; J + +) {                if(j>i) Dp[i][j] =0; Else if(j==i) Dp[i][j] =1; ElseDP[I][J] = dp[i-1][j-1] + dp[i-J]            [j]; }} B=Dp[n][k]; //divide I into odd numbers of no more than J        intOdd = n&1? n:n-1;  for(intI=1; i<=n; i++) {dp[i][1] =1;  for(intj=3; j<=odd; j+=2) {                if(j>i) {if(i&1) Dp[i][j] =Dp[i][i]; ElseDP[I][J] = dp[i][i-1]; }                Else if(j==i) Dp[i][j] = dp[i][i-2] +1; ElseDP[I][J] = dp[i][j-2] + dp[i-J]            [j]; }} D=Dp[n][odd]; //Dividing I into the number of different positive integers         for(intI=1; i<=n; i++) {             for(intj=1; j<=n; J + +) {                if(j>i) Dp[i][j] =Dp[i][i]; Else if(j = i) dp[i][j] = dp[i][i-1] +1; ElseDP[I][J] = dp[i][j-1] + dp[i-j][j-1]; }} e=Dp[n][n]; printf ("%d\n%d\n%d\n%d\n%d\n\n", A, B, C, D, E); Memset (DP,0,sizeof(DP)); }        return 0;}

Nyoj 571--Various Division numbers