大数乘除（hdoj1023)

  通过这题深深的理解了算法的重要性。

  深深理解了算法没有绝对最优，只有相对，要针对具体情况设计相应的算法，这才是王道！

之前做死的写大数*大数，大数/大数，好不容易写出来，TLE，改成10^10进制，在自己机器上运行居然比10进制的还慢！后来发现是大数/大数出的问题，方法是用减法模拟除法，然后就悲剧了如果是像10^10/2这样很小的数，就这一个数都会TLE.

  后来仔细观察其实没必要用大数*大数，大数/大数，只要大数/（小数）就行。然后10^10进制果断0MS过了。因为c[i]=c[i-1]*(4*i-2)/(i+1)

TLE的代码：

#include<iostream>using namespace std ;#define N 60//every element in array is between 0 and 9//reverse the data to restore,a0,a1,...,an-1void mul(int *a,int *b,int *c,int lenA,int lenB,int &lenC){int i;lenC=lenA+lenB-1;//[lena+lenb-1,lena+lenb]for(i=0;i<=lenC;i++)c[i]=0;//<= not < for the jugde lenD++;for(i=0;i<lenA;i++){for(int j=0;j<lenB;j++)c[i+j]+=a[i]*b[j];}for(i=0;i<lenC;i++){if(c[i]>9){c[i+1]+=(c[i]/10);c[i]%=10;}}i=lenC;while(i>0&&c[i]==0)i--;lenC=i+1;}//if a>=b return true;bool compare(int *a,int *b,int lenA,int lenB){if(lenA>lenB&&a[lenA-1]!=0)return true;//zero serieselse if(lenA<lenB)return false;else{int i;for(i=lenA-1;i>-1;i--){if(a[i]>b[i])return true;else if(a[i]<b[i])return false;}return true;}}//a-b*10^(lenA-lenB)//every element in array is between 0 and 9//we assume a>=b , both a and b are positive//reverse the data to restore,a0,a1,...,an-1void sub(int *a,int *b,int *c,int lenA,int lenB,int &lenC){fill(c,c+lenA,0);int i;for(i=0;i<lenB;i++){c[i]+=(a[i]-b[i]);if(c[i]<0)c[i]+=10,c[i+1]--;}for(i=lenB;i<lenA;i++)c[i]+=a[i];i=lenA-1;while(i>=1&&c[i]==0)i--;//lenC=i+1;}//we assume a>=b ,both a and b are positive//at last,the rest of a/b is in a//we assume a>=b , both a and b are positive//reverse the data to restore,a0,a1,...,an-1void div(int *a,int *b,int *c,int lenA,int lenB,int &lenC){//int &lenA用于求余数,注意a数组值是变化的！lenC=lenA-lenB+1;if(a[lenA-1]<b[lenB-1])lenC--;if(lenC<=0){lenC=1;c[0]=0;}else{fill(c,c+lenC,0);int tmp[N],len;int i;for(i=lenC-1;i>-1;i--){while(compare(a+i,b,lenA-i,lenB)){//particular position handlec[i]++;sub(a+i,b,tmp,lenA-i,lenB,len);memcpy(a+i,tmp,len*sizeof(int));lenA=i;if(len!=1||tmp[0]!=0)lenA+=len;}}}}void itoi(int source,int *des,int &len){len=0;while(source){des[len++]=source%10;source/=10;}}int c[101][N],cLen[101];void catalan(int n){c[0][0]=1;cLen[0]=1;int convert[N],conLen,t[N],tLen;for(int i=1;i<=n;i++){itoi(4*i-2,convert,conLen);mul(convert,c[i-1],t,conLen,cLen[i-1],tLen);itoi(i+1,convert,conLen);div(t,convert,c[i],tLen,conLen,cLen[i]);}}int main(){catalan(100);int n;while(scanf("%d",&n)!=EOF){int i;for(i=cLen[n]-1;i>-1;i--)cout<<c[n][i];cout<<endl;}return 0;}

0msAC的代码

#include<iostream>using namespace std ;#define LL __int64#define RADIX 10000000000//radix=LL_MAX,10位一组，LL*LL<__LL64,用LL存储#define N 10//reverse the data to restore,a0,a1,...,an-1//a*=bvoid mul(LL *a,int b,int &lenA){int i;for(i=0;i<lenA;i++)a[i]*=b;for(i=0;i<lenA;i++){if(a[i]>=RADIX){a[i+1]+=(a[i]/RADIX);a[i]%=RADIX;}}if(a[lenA])lenA++;for(i=lenA-1;i>0&&(!a[i]);i--);lenA=i+1;}void div(LL *a,int b,int &lenA){int i;for(i=lenA-1;i>-1;i--){if(i>0)a[i-1]+=((a[i]%b)*RADIX);a[i]/=b;}for(i=lenA-1;i>0&&(!a[i]);i--);lenA=i+1;}LL c[101][N];int cLen[101];void catalan(LL n){c[0][0]=1;cLen[0]=1;for(LL i=1;i<=n;i++){memcpy(c[i],c[i-1],cLen[i-1]*sizeof(LL));cLen[i]=cLen[i-1];mul(c[i],4*i-2,cLen[i]);div(c[i],i+1,cLen[i]);}}void output(LL *a,int lenA){LL i;LL t[N];memcpy(t,a,lenA*sizeof(LL));printf("%I64d",a[lenA-1]);for(i=lenA-2;i>-1;i--){LL cnt=10;while(t[i]){t[i]/=10;cnt--;}while(cnt--)printf("0");printf("%I64d",a[i]);}}int main(){catalan(100);LL n;while(scanf("%d",&n)!=EOF){output(c[n],cLen[n]);cout<<endl;}return 0;}