Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15663 Accepted Submission(s): 11042
"The second problem is, given an positive integer N, we define an equation like this:
N=a[1]+a[2]+a[3]+...+a[m];
a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"
题意:给定一个整数,将其写成若干个整数的和,求有多少种情况。
可以直接考虑各个加数递减的情况,避免重复。依旧是DP的思想。
将整数n划分成两个整数之和,n+0,(n-1)+1,(n-2)+2,……,(n-n/2)+(n/2)。再将加号前面的加数继续如此划分,注意:继续划分时,后一个加数至少为上一级的后一个加数,以确保加数列的递减性。用a[n][1]表示将n表示成(n-1)+1的情况数,则可知a[n][1]=a[n-1][0]+a[n-1][1]+a[n-1][2]+……+a[n-1][(n-1)/2]+1。
a[i][j]=a[i-1][0]+a[i-1][j]+a[i-1][j+1]+……+a[i-1][(i-1)/2]。把a[i][0]都初始化为1,表示不划分时就一种情况。最后,将所有a[i][j]都加到a[i][0]上,表示i的所有划分情况数。
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int a[125][70];//a[i][j]表示和为i 且 两个加数的后一个为j的情况数量(j<=i/2)
void biao()
{
for(int i = 0; i < 125; i++)
{
a[i][0]=1;
for(int j = 1; j <= i / 2; j++)
{
for(int k = j; k <= (i - j) / 2; k++)
a[i][j] += a[i-j][k];
a[i][j]++;
a[i][0] += a[i][j];
}
}
}
int main()
{
//freopen("in.txt","r",stdin);
biao();
int n;
while (scanf("%d",&n)!=EOF)
{
printf("%d\n",a[n][0]);
}
}