以前从来没写过,因为过于复杂,现在回头在看,也就那么回事。看来在复杂的东西只要坐下来尽心的去写,就一定能写出来。附上我写的高精算法,望高手指教。(可能冗余的代码很多,还没来的急优化)
#include<iostream>
#include<string.h>
typedef struct{
int arr[255];
int s;
}martrix;
martrix chengfa(martrix A,martrix B){
int cw,jw,a[255],b[255],c[255],d[255],ds,T,cc,ca,cb,i,j,k,p;
martrix shu;
ca=A.s;cb=B.s;
for(i=1;i<=A.s;i++)
a[i]=A.arr[i];
for(i=1;i<=B.s;i++)
b[i]=B.arr[i];
memset(c,0,sizeof(c));
memset(d,0,sizeof(d));
cw=1;
for(i=1;i<=cb;i++){
jw=0;
for(j=1;j<=ca;j++){
d[j]=(a[j]*b[i]+jw)%10000;
jw=(a[j]*b[i]+jw)/10000;
}
if(jw!=0){
d[j]=jw;
ds=j;
}
else ds=j-1;
jw=0;
for(k=cw,p=1;p<=ds;p++,k++){
T=d[p]+c[k]+jw;
c[k]=T%10000;
jw=T/10000;
}
if(jw!=0){
c[k]=jw;
cc=k;
}
else cc=k-1;
cw++;
}
shu.s=cc;
for(i=1;i<=cc;i++)
shu.arr[i]=c[i];
return shu;
}
int main(){
char s1[255],s2[255];
int n,i,j,k,L,xs,f;
memset(s1,'\0',sizeof(s1));
memset(s2,'\0',sizeof(s2));
while(scanf("%s%d",s1,&n)!=EOF){
L=strlen(s1);
for(i=0;i<L;i++){
if(s1[i]=='.')
break;
}
if(i>=L){
s1[L++]='.';
s1[L]='\0';
}
for(i=0;i<L;i++){
if(s1[i]!='0'||s1[i]=='.')
break;
}
for(j=L-1;j>=0;j--){
if(s1[j]!='0'||s1[j]=='.')
break;
}
xs=0;f=1;
for(k=0;i<=j;i++){
if(s1[i]!='.'&&f){
xs++;
}
else
f=0;
s2[k++]=s1[i];
}
s2[k]='\0';
xs=strlen(s2)-xs-1;
xs=xs*n;
int a[255],t[255],c;
martrix g,h;
L=strlen(s2);
for(i=L-1,k=1;i>=0;i--){
if(s2[i]!='.'){
a[k++]=s2[i]-48;
}
}
for(k=1,i=1;(i+4)<=L-1;i+=4)
t[k++]=a[i]+a[i+1]*10+a[i+2]*100+a[i+3]*1000;
c=L-i;
if(c!=0){
if(c==1) t[k]=a[i];
if(c==2) t[k]=a[i]+a[i+1]*10;
if(c==3) t[k]=a[i]+a[i+1]*10+a[i+2]*100;
if(c==4) t[k]=a[i]+a[i+1]*10+a[i+2]*100+a[i+3]*1000;
g.s=k;
}
else g.s=k-1;
for(i=1;i<=g.s;i++)
g.arr[i]=t[i];
h=g;
for(i=1;i<n;i++)
g=chengfa(g,h);
char ch[255]={'\0'};
sprintf(ch,"%d",g.arr[g.s]);
for(i=g.s-1;i>=1;i--){
if(g.arr[i]<10) sprintf(ch,"%s000%d",ch,g.arr[i]);
else if(g.arr[i]<100) sprintf(ch,"%s00%d",ch,g.arr[i]);
else if(g.arr[i]<1000) sprintf(ch,"%s0%d",ch,g.arr[i]);
else sprintf(ch,"%s%d",ch,g.arr[i]);
}
if((int)strlen(ch)>xs){
for(k=0;k<(int)strlen(ch);k++){
if((int)strlen(ch)-k==xs)
printf(".");
printf("%c",ch[k]);
}
}
else{
printf(".");
for(i=1;i<=xs-(int)strlen(ch);i++)
printf("0");
for(i=0;i<(int)strlen(ch);i++)
printf("%c",ch[i]);
}
printf("\n");
memset(s1,'\0',sizeof(s1));
memset(s2,'\0',sizeof(s2));
}
}
转载于:https://www.cnblogs.com/saintqdd/archive/2007/08/01/838709.html