Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8447 Accepted Submission(s): 6009
Problem Description
"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.
"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!"
"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!"
Input
The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.
Output
For each test case, you have to output a line contains an integer P which indicate the different equations you have found.
Sample Input
4
10
20
Sample Output
5
42
627
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 int main(void){ 6 #ifndef ONLINE_JUDGE 7 freopen("1028.in", "r", stdin); 8 #endif 9 int dp[121][121]; 10 memset(dp, 0, sizeof(dp)); 11 for (int i = 0; i < 121; ++i){ 12 dp[i][1] = dp[1][i] = 1; 13 } 14 for (int i = 2; i < 121; ++i){ 15 for (int j = 2; j < 121; ++j){ 16 if (i < j) dp[i][j] = dp[i][i]; 17 else if (i == j) dp[i][j] = 1 + dp[i][j-1]; 18 else dp[i][j] = dp[i][j-1] + dp[i-j][j]; 19 } 20 } 21 int n; 22 while (~scanf("%d", &n)){ 23 printf("%d\n", dp[n][n]); 24 } 25 return 0; 26 }
突然感觉动态规划特别有意思……
这道题目也可以用母函数做，可惜看了一下，没有看懂……
dp[i][j] 表示把整数i用不大于j的正整数之和表示.
分三种情况：
当ｉ》ｊ的时候，如果选ｊ，那么有dp[i-j][j]种；如果不选ｊ，那么有ｄｐ[ｉ][ｊ-1]种；
当ｉ==ｊ的时候，如果选ｊ，那么有1种；如果不选ｊ，那么又ｄｐ[ｉ][ｊ-1]种；
当ｉ《ｊ的时候，则ｄｐ[i][j] = dp[i][i]；
然后，显然，当ｉ或者ｊ其中一个等于1的时候，结果都是1.
求ｎ的划分，就是求把整数ｎ用不大于ｎ的正整数之和表示，有几种方法，就是ｄｐ[n][n]。