1025: coin statistics

Description

Assume that a pile of n coins, consisting of 1, 2, and 5, have a total face value of M, evaluate the total number of possible combinations (the number of coins with a certain nominal value can be 0 ).

Input Format

The first line of the input data has a positive integer T, indicating that there are T groups of test data. In the next t row, each row has two numbers n, m, n, and M.

Output

For each group of test data, please output the number of possible combinations, each group of output occupies one line.

Sample Input

2
3 5
4 8

Sample output

1
2

The idea of this question is similar to that of chicken, rabbit, and cage, so it is not difficult to think of using a few for loops to perform exhaustive actions on possible values. Below is an algorithm I have written, which is slightly optimized in exhaustive actions.

 1 #include <stdio.h> 2 int main(void) 3 { 4     int n,m; 5     int time; 6      7     scanf("%d",&time); 8     while(time--) 9     {    10         11         int count=0;12         scanf("%d %d",&n,&m);13         int i,j,k,total;14         15         for(i=0;i<=(m/5);i++)16         {17             18             for(j=0;j<=(m/2);j++)19                 {20                     k=n-j-i;                21                     total=k*1+j*2+i*5;22                     if(total==m)23                             count++;24                 }25                 26         }27         printf("%d\n",count);28     }29     return 0;30 }

There are still errors after submission, and no where is found yet. The following is an official algorithm with some optimizations.

 1 #include<stdio.h> 2  3 int main() 4 { 5         int t,n,m,c1,c2,c5,k; 6         scanf("%d",&t); 7         while(t--) 8         { 9                 scanf("%d%d",&n,&m);10                 k=0;11                 for(c5=0;5*c5<=m;c5++)12                         for(c2=0;2*c2+5*c5<=m;c2++)13                         {14                                 c1=m-5*c5-2*c2;15                                 if(c1+c2+c5==n)16                                         k++;17                         }18                 printf("%d\n",k);19         }20         return 0;21 }

It is also worth mentioning that this question is similar to the algorithm of the 1023-hookay black store.

In addition, it was modified according to the official website. Strange.

 1 #include <stdio.h> 2 int main(void) 3 { 4     int n,m; 5     int time; 6      7     scanf("%d",&time); 8     while(time--) 9     {    10         11         int count=0;12         scanf("%d %d",&n,&m);13         int i,j,k,total;14         15         for(i=0;5*i<=m;i++)16         {17             18             for(j=0;2*j<=m;j++)19                 {20                     k=n-j-i;                21                     total=k*1+j*2+i*5;22                     if(total==m)23                             count++;24                 }25                 26         }27         printf("%d\n",count);28     }29     return 0;30 }

Finally, I finally found the problem. k = n-I-j; because there is no limit on the initial j of I, K may be a negative value.

The following code is AC

 1 #include <stdio.h> 2 int main(void) 3 { 4     int n,m; 5     int time; 6      7     scanf("%d",&time); 8     while(time--) 9     {    10         11         int count=0;12         scanf("%d %d",&n,&m);13         int i,j,k,total;14         15         for(i=0;5*i<=m;i++)16         {17             18             for(j=0;2*j<=m;j++)19                 {20                     k=n-j-i;                21                     total=k*1+j*2+i*5;22                     if(total==m&&k>=0)23                     {24                         25                         count++;26                     }27                             28                 }29                 30         }31         printf("%d\n",count);32     }33     return 0;34 }

 

1025: coin statistics