HDU 1131 Count the Trees large number calculation, hdu1131

HDU 1131 Count the Trees large number calculation, hdu1131

The question is to give a number, and then use 1 to this number as the node of the binary tree. There are several methods to form the binary tree. This question is to use the catlan number to calculate the number, and then because there is a number, different numbers correspond to different methods, so the result is the catlan number multiplication method of this number. Because the numbers are relatively large, we need to use a high-precision method, that is, using character arrays. I have written three functions, one for addition, one for multiplication, And the last for table creation, when the table is typed, you only need to judge the number of input items and directly output the data. below is my code. The first time I write this type of big data processing with high precision, it may seem complicated =

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char ans[105][1005];char ads[1005],mus[1005];char jc[105][1005];char res[105][1005];int multiply(char* a1,char* b1){    memset(mus,0,sizeof(mus));    int i,j,k;    int len,len1,len2;    len1=strlen(a1);    len2=strlen(b1);    int mut=0,t;    bool va[1005];    int s[1005],adt[1005];    char a[1005],b[1005];    memset(s,0,sizeof(s));    memset(adt,0,sizeof(adt));    memset(va,0,sizeof(va));    for(i=len1-1,j=0;i>=0;i--)    {        a[i]=a1[j];        j++;    }    for(i=len2-1,j=0;i>=0;i--)    {        b[i]=b1[j];        j++;    }    for(i=0;i<len1;i++)    {        mut=0;        for(j=0;j<len2;j++)        {            t=(a[i]-'0')*(b[j]-'0')+mut;            mut=0;            if(t>=10)            {                mut=t/10;                t=t%10;            }            s[i+j]=t+s[i+j]+adt[i+j];            va[i+j]=1;            adt[i+j]=0;            if(s[i+j]>=10)            {                if(!va[i+j+1])                    adt[i+j+1]=s[i+j]/10+adt[i+j+1];                else                    adt[i+j+1]=s[i+j]/10;                s[i+j]=s[i+j]%10;            }        }        s[i+j]=mut;    }    s[i+j-1]=mut+adt[i+j-1];    if(s[i+j-1]!=0)        k=i+j-1;    else        k=i+j-2;    for(i=k,j=0;i>=0;i--)    {        mus[i]=(s[j]+'0');        j++;    }    return 0;}int additive(char* a,char* b){    memset(ads,0,sizeof(ads));    int len,len1,len2;    int i;    int ad[1005];    len1=strlen(a);    len2=strlen(b);    if(len1==len2)    {        len=len1;    }    else if(len1>len2)    {        len=len1;        for(i=len;i>=len-len2;i--)        {            b[i]=b[i-len+len2];        }        for(i=len-len2-1;i>=0;i--)        {            b[i]='0';        }    }    else if(len1<len2)    {        len=len2;        for(i=len;i>=len-len1;i--)        {            a[i]=a[i-len+len1];        }        for(i=len-len1-1;i>=0;i--)        {            a[i]='0';        }    }    int t=0;    for(i=len-1;i>=0;i--)    {        ad[i]=(a[i]-'0')+(b[i]-'0')+t;        t=0;        if(ad[i]>=10)        {            t++;            ad[i]=ad[i]-10;            ads[i]=ad[i]+'0';        }        else        {            ads[i]=ad[i]+'0';        }    }    if(t==1)    {        for(i=len;i>=0;i--)        {            ads[i]=ads[i-1];        }        ads[0]='1';    }    return 0;}int excel(){    ans[0][0]='1';    ans[1][0]='1';    char sum[105];    int n;    int i,j;    char t[5];    memset(sum,0,sizeof(sum));    for(i=2;i<=100;i++)    {        for(j=i;j>0;j--)        {            multiply(ans[i-j],ans[j-1]);            additive(mus,sum);            strcpy(sum,ads);        }        strcpy(ans[i],sum);        memset(sum,0,sizeof(sum));    }    jc[1][0]='1';    for(i=2;i<100;i++)    {        memset(t,0,sizeof(t));        if(i>=10)        {            t[0]=i/10+'0';            t[1]=i%10+'0';        }        else        {            t[0]=i+'0';        }        multiply(jc[i-1],t);        strcpy(jc[i],mus);    }    multiply(jc[99],"100");    strcpy(jc[100],mus);    for(i=1;i<=100;i++)    {        multiply(ans[i],jc[i]);        strcpy(res[i],mus);        //cout<<"res["<<i<<"]="<<res[i]<<endl;    }    return 0;}int main(){    int n;    excel();    while(scanf("%d",&n)!=EOF)    {        if(n==0)            break;        cout<<res[n]<<endl;    }    return 0;}