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。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

const int MAX = 1000+5;
typedef  __int64 INT;
char dp[105][10000];

void sum(char* s1,char* s2) {
    int len1 = strlen(s1);
    int len2 = strlen(s2);
    int a1[MAX],a2[MAX];
    memset(a1,0,sizeof(a1));
    memset(a2,0,sizeof(a2));
    for(int i = 0;i<len1;++i)
    a1[i] = s1[i] - '0';
    for(int i = 0;i<len2;++i)
    a2[i] = s2[i] - '0';
    int m = max(len1,len2);
    for(int i = 0;i<m;++i) {
        a1[i] += a2[i];
        a1[i+1] += (a1[i]/10);
        a1[i]%=10;
    }
    if(a1[m] != 0)
    m++;
    strcpy(s1,"");
    for(int i = 0;i<m;++i)
    s1[i] = a1[i]+'0';
    s1[m] = NULL;
}

char *mult(char* s1,char* s2) {
    char s4[MAX],*s3 = new char[MAX];
    strcpy(s3,"0");
    int len1 = strlen(s1);
    int len2 = strlen(s2);
    for(int i = 0;i<len2;++i) {
        for(int j = 0;j<i;++j)
        s4[j] = '0';
        int a[MAX];
        memset(a,0,sizeof(a));
        for(int j = 0;j<len1;++j)
        a[j] = s1[j] - '0';
        int d = s2[i]-'0';
        for(int j = 0;j<len1;++j)
        a[j] *= d;
        for(int j = 0;j<len1;++j) {
            a[j+1] += (a[j]/10);
            a[j] %= 10;
        }
        int m = len1;
        if(a[m] != 0)
        m++;
        for(int j = 0;j<m;++j)
        s4[j+i] = a[j]+'0';
        s4[m+i] = NULL;
        sum(s3,s4);
    }
    return s3;
}
void dabiao() {
    strcpy(dp[1],"0");
    strcpy(dp[2],"1");
    strcpy(dp[3],"1");
    for(int i = 4;i <= 102;++i)
    {
        char str3[1000] = "0";
        for(int j = 2;j <= i - 1;++j)
        sum(str3,mult(dp[j],dp[i-j+1]));
        strcpy(dp[i],str3);
    }
}

int main( ) {
    int n;
    dabiao();
    while(scanf("%d",&n)!=EOF)
    {
        int len = strlen(dp[n+2]);
        for(int i = len-1;i>=0;--i)
        printf("%c",dp[n+2][i]);
        puts("");
    }
    return 0;
}