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 ≤ 10 ¹²)

 

 

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：

 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       else

59       {

60       memset(aa,'\0',sizeof(aa));

61       memset(bb,'\0',sizeof(bb));

62       memset(ans,0,sizeof(ans));

63       //,(8*n-7)*n+1

64       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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