HDU ---- (5047) sawtooth (multiplication of large numbers + mathematical derivation)

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 ≤0 ≤ 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 input212

 

Sample outputcase #1: 2 case #2: 19

 

Source 2014 ACM/ICPC Asia Regional Shanghai Online in fact, the question has clearly told us that it is derived from the cable fragment .... for the line distribution plane 0 11 1 + 1 2 1 + 1 + 23 1 + 1 + 2 + 34 1 + 1 + 2 + 3 + 4 ............ N 1 + n (n + 1)/2; then for an M model, we can regard it as a combination of four line segments, and this formula becomes: 4N * (4n + 1)/2 + 1 ----> obviously, the answer is more than enough. I am 0 11 11 11 2 92 37 19 9*2 ...... push to get: 4N * (4n + 1)/2 + 1-8 * n ----> 8n ^ 2-7n + 1 code:

 1 #include<cstdio> 2 #include<cstring> 3 char aa[50],bb[50]; 4 int ans[50]; 5 int mul( char *a, char *b, int temp[]) 6 { 7  8     int i,j,la,lb,l; 9     la=strlen(a);10     lb=strlen(b);11 12     for ( i=0;i<la+lb;i++ )13         temp[i]=0;14     for ( i=0;i<=la-1;i++ ) {15           l=i;16         for ( j=0;j<=lb-1;j++ ) {17             temp[l]=(b[j]-‘0‘)*(a[i]-‘0‘)+temp[l];18             l++;19         }20     }21     while ( temp[l]==0 )22         l--;23     for ( i=0;i<=l;i++ ) {24         temp[i+1]+=temp[i]/10;25         temp[i]=temp[i]%10;26     }27     if ( temp[l+1]!=0 )28         l++;29 30     while ( temp[l]/10!=0 ) {31         temp[l+1]+=temp[l]/10;32         temp[l]=temp[l]%10;33         l++;34     }35     if ( temp[l]==0 )36         l--;37     return l;38 }39 void cal(__int64 a,char *str)40 {41     int i=0;42     while(a>0)43     {44      str[i++]=(a%10)+‘0‘;45      a/=10;46     }47 }48 int main()49 {50     int cas;51     __int64 n;52     scanf("%d",&cas);53     for(int i=1;i<=cas;i++)54     {55       scanf("%I64d",&n);56       printf("Case #%d: ",i);57       if(n==0)printf("1\n");58       else59       {60       memset(aa,‘\0‘,sizeof(aa));61       memset(bb,‘\0‘,sizeof(bb));62       memset(ans,0,sizeof(ans));63       //,(8*n-7)*n+164       cal(8*n-7,aa);65       cal(n,bb);66       int len=mul(aa,bb,ans);67        ans[0]++;68        int c=0;69      for(int j=0;j<=len;j++)70      {71          ans[j]+=c;72        if(ans[j]>9)73         {74           c=ans[j]/10;75           ans[j]%=10;76         }77      }78       if(c>0)79         printf("%d",c);80       for(int j=len;j>=0;j--)81         printf("%d",ans[j]);82     printf("\n");83     }84     }85  return 0;86 }

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HDU ---- (5047) sawtooth (multiplication of large numbers + mathematical derivation)