HDU 1023 Train Problem II
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1023
解题报告:就是求第n个卡特兰数是多少，不过这个n的范围有点大1到100，所以还要用到高精度，其实这题也就是纯粹的高精度加法跟乘法结合起来。
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 using namespace std; 5 6 const int MAX = 1000+5; 7 typedef __int64 INT; 8 char dp[105][10000]; 9 10 void sum(char* s1,char* s2) { 11 int len1 = strlen(s1); 12 int len2 = strlen(s2); 13 int a1[MAX],a2[MAX]; 14 memset(a1,0,sizeof(a1)); 15 memset(a2,0,sizeof(a2)); 16 for(int i = 0;i<len1;++i) 17 a1[i] = s1[i] - '0'; 18 for(int i = 0;i<len2;++i) 19 a2[i] = s2[i] - '0'; 20 int m = max(len1,len2); 21 for(int i = 0;i<m;++i) { 22 a1[i] += a2[i]; 23 a1[i+1] += (a1[i]/10); 24 a1[i]%=10; 25 } 26 if(a1[m] != 0) 27 m++; 28 strcpy(s1,""); 29 for(int i = 0;i<m;++i) 30 s1[i] = a1[i]+'0'; 31 s1[m] = NULL; 32 } 33 34 char *mult(char* s1,char* s2) { 35 char s4[MAX],*s3 = new char[MAX]; 36 strcpy(s3,"0"); 37 int len1 = strlen(s1); 38 int len2 = strlen(s2); 39 for(int i = 0;i<len2;++i) { 40 for(int j = 0;j<i;++j) 41 s4[j] = '0'; 42 int a[MAX]; 43 memset(a,0,sizeof(a)); 44 for(int j = 0;j<len1;++j) 45 a[j] = s1[j] - '0'; 46 int d = s2[i]-'0'; 47 for(int j = 0;j<len1;++j) 48 a[j] *= d; 49 for(int j = 0;j<len1;++j) { 50 a[j+1] += (a[j]/10); 51 a[j] %= 10; 52 } 53 int m = len1; 54 if(a[m] != 0) 55 m++; 56 for(int j = 0;j<m;++j) 57 s4[j+i] = a[j]+'0'; 58 s4[m+i] = NULL; 59 sum(s3,s4); 60 } 61 return s3; 62 } 63 void dabiao() { 64 strcpy(dp[1],"0"); 65 strcpy(dp[2],"1"); 66 strcpy(dp[3],"1"); 67 for(int i = 4;i <= 102;++i) 68 { 69 char str3[1000] = "0"; 70 for(int j = 2;j <= i - 1;++j) 71 sum(str3,mult(dp[j],dp[i-j+1])); 72 strcpy(dp[i],str3); 73 } 74 } 75 76 int main( ) { 77 int n; 78 dabiao(); 79 while(scanf("%d",&n)!=EOF) 80 { 81 int len = strlen(dp[n+2]); 82 for(int i = len-1;i>=0;--i) 83 printf("%c",dp[n+2][i]); 84 puts(""); 85 } 86 return 0; 87 }
