母函数

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23902    Accepted Submission(s): 16627

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!"

 

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

 

Author

Ignatius.L

 

Recommend

We have carefully selected several similar problems for you:  1171 1085 1398 2152 1709 

#include <iostream>
#include <string>
#include <algorithm>
#include <string.h>
#include <cstring>
#include <math.h>
#include <stdio.h>

#define mem(a) memset(a,0,sizeof(a))
#define maxn 150
#define ll long long
using namespace std;
int a[maxn],b[maxn];

int main(){
    ios::sync_with_stdio(0);
    int n;
    while(scanf("%d",&n)!=EOF){
    for(int i=0;i<=n;i++){
        a[i]=1;
        b[i]=0;
    }

     for(int i=2;i<=n;i++){
        for(int j=0;j<=n;j++)
            for(int k=0;k+j<=n;k+=i)
                b[k+j]+=a[j];
        for(int i=0;i<=n;i++){
        a[i]=b[i];
        b[i]=0;
        }
     }
      printf("%d
",a[n]);
    }
    return 0;
}

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13374    Accepted Submission(s): 9194

Problem Description

People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ..., and 289-credit coins, are available in Silverland. 
There are four combinations of coins to pay ten credits: 

ten 1-credit coins,
one 4-credit coin and six 1-credit coins,
two 4-credit coins and two 1-credit coins, and
one 9-credit coin and one 1-credit coin. 

Your mission is to count the number of ways to pay a given amount using coins of Silverland.

 

Input

The input consists of lines each containing an integer meaning an amount to be paid, followed by a line containing a zero. You may assume that all the amounts are positive and less than 300.

 

Output

For each of the given amount, one line containing a single integer representing the number of combinations of coins should be output. No other characters should appear in the output. 

 

Sample Input

2 10 30 0

 

Sample Output

1 4 27

 

Source

Asia 1999, Kyoto (Japan)

 

#include <iostream>
#include <string>
#include <algorithm>
#include <string.h>
#include <cstring>
#include <math.h>
#include <stdio.h>

#define mem(a) memset(a,0,sizeof(a))
#define maxn 350
#define ll long long
using namespace std;
//int money[]={0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289};
int a[maxn],b[maxn];

int main(){
    ios::sync_with_stdio(0);
    int n;
    while(scanf("%d",&n)&&n){
        for(int i=0;i<=n;i++){
            a[i]=1;
            b[i]=0;
        }
        for(int i=2;i*i<=n;i++){
              for(int j=0;j<=n;j++)
                for(int k=0;k+j<=n;k+=i*i)
                    b[k+j]+=a[j];

            for(int i=0;i<=n;i++){
                a[i]=b[i];
                b[i]=0;
            }
        }
        printf("%d
",a[n]);
    }
    return 0;
}

 

Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24244    Accepted Submission(s): 10777

Problem Description

We all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China! 
“Oh, God! How terrible! ”



Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up! 
Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem?
“Given some Chinese Coins (硬币) (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.”
You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush!

 

Input

Input contains multiple test cases. Each test case contains 3 positive integers num_1, num_2 and num_5 (0<=num_i<=1000). A test case containing 0 0 0 terminates the input and this test case is not to be processed.

 

Output

Output the minimum positive value that one cannot pay with given coins, one line for one case.

 

Sample Input

1 1 3 0 0 0

 

Sample Output

4

 

Author

lcy

#include <iostream>
#include <string>
#include <algorithm>
#include <string.h>
#include <cstring>
#include <math.h>
#include <stdio.h>
#include <cstdio>

#define mem(a) memset(a,0,sizeof(a))
#define maxn 8010
#define BASE 10000
#define ll long long

using namespace std;
/*
 int a[20];
    sort(a, a+20, [](int a, int b){return a > b;})
*/

int a[maxn],b[maxn];
int num[5];

int main(){
    ios::sync_with_stdio(0);
        while(scanf("%d%d%d",&num[0],&num[1],&num[2])){
            if(num[0]==0&&num[1]==0&&num[2]==0) return 0;
            int maxx=num[0]*1+num[1]*2+num[2]*5;
            for(int i=0;i<=maxx;i++){
                a[i]=1;
                b[i]=0;
            }
            for(int i=0;i<=num[0];++i)
                a[i]=1;
            for(int i=0;i<=num[0];++i)
                for(int j=0;j<=num[1]*2;j+=2)
                    b[j+i]+=a[i];
            for(int i=0;i<=num[1]*2+num[0]*1;i++){
                a[i]=b[i];
                b[i]=0;
            }
            for(int i=0;i<=num[0]*1+num[1]*2;i++)
                for(int j=0;j<=num[2]*5;j+=5)
                    b[j+i]+=a[i];
            for(int i=0;i<=maxx;i++){
                a[i]=b[i];
                b[i]=0;
            }
            int i;
            for(i=0;i<=maxx;i++){
                if(a[i]==0){
                    printf("%d
",i);
                    break;
                }
            }
            if(i==maxx+1)
                printf("%d
",i);
        }
    return 0;
}