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：
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 }
View Code
编程是一种快乐，享受代码带给我的乐趣！！！
