hdu----(5047)Sawtooth(大数相乘+数学推导)

Sawtooth

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 422    Accepted Submission(s): 134

Problem Description

Think about a plane:

● One straight line can divide a plane into two regions.
● Two lines can divide a plane into at most four regions.
● Three lines can divide a plane into at most seven regions.
● And so on...

Now we have some figure constructed with two parallel rays in the same direction, joined by two straight segments. It looks like a character “M”. You are given N such “M”s. What is the maximum number of regions that these “M”s can divide a plane ?

 

Input

The first line of the input is T (1 ≤ T ≤ 100000), which stands for the number of test cases you need to solve.

Each case contains one single non-negative integer, indicating number of “M”s. (0 ≤ N ≤ 10¹²)

 

Output

For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then an integer that is the maximum number of regions N the “M” figures can divide.

 

Sample Input

2 1 2

 

Sample Output

Case #1: 2 Case #2: 19

 

Source

2014 ACM/ICPC Asia Regional Shanghai Online

 

 

其实题目已经很清楚的告知我们是有线条分平面引申而来的了....

对于线条分平面

0  1

1  1 +1

2  1+1 +2

3 1+1 +2+3

4 1+1 +2+3+4

............

n   1+n(n+1)/2;

那么对于一个m型号的模型，其实我们可以将其视其为四条线段组合而成，这样这个公式就变为：

 4n*(4n+1)/2 +1  ---->显然得到的答案有余坠，我

0  1

1   11    2       9

2   37    19     9*2

......

推到得到：

 4n*(4n+1)/2  +1 -8*n----> 8n^2-7n+1

代码：

#include<cstdio>
#include<cstring>
char aa[50],bb[50];
int ans[50];
int mul( char *a, char *b, int temp[])
{

    int i,j,la,lb,l;
    la=strlen(a);
    lb=strlen(b);

    for ( i=0;i<la+lb;i++ )
        temp[i]=0;
    for ( i=0;i<=la-1;i++ ) {
          l=i;
        for ( j=0;j<=lb-1;j++ ) {
            temp[l]=(b[j]-'0')*(a[i]-'0')+temp[l];
            l++;
        }
    }
    while ( temp[l]==0 )
        l--;
    for ( i=0;i<=l;i++ ) {
        temp[i+1]+=temp[i]/10;
        temp[i]=temp[i]%10;
    }
    if ( temp[l+1]!=0 )
        l++;

    while ( temp[l]/10!=0 ) {
        temp[l+1]+=temp[l]/10;
        temp[l]=temp[l]%10;
        l++;
    }
    if ( temp[l]==0 )
        l--;
    return l;
}
void cal(__int64 a,char *str)
{
    int i=0;
    while(a>0)
    {
     str[i++]=(a%10)+'0';
     a/=10;
    }
}
int main()
{
    int cas;
    __int64 n;
    scanf("%d",&cas);
    for(int i=1;i<=cas;i++)
    {
      scanf("%I64d",&n);
      printf("Case #%d: ",i);
      if(n==0)printf("1
");
      else
      {
      memset(aa,'',sizeof(aa));
      memset(bb,'',sizeof(bb));
      memset(ans,0,sizeof(ans));
      //,(8*n-7)*n+1
      cal(8*n-7,aa);
      cal(n,bb);
      int len=mul(aa,bb,ans);
       ans[0]++;
       int c=0;
     for(int j=0;j<=len;j++)
     {
         ans[j]+=c;
       if(ans[j]>9)
        {
          c=ans[j]/10;
          ans[j]%=10;
        }
     }
      if(c>0)
        printf("%d",c);
      for(int j=len;j>=0;j--)
        printf("%d",ans[j]);
    printf("
");
    }
    }
 return 0;
}