POJ 1001 Exponentiation 求高精度幂

标签：c++   iostream   poj   编程   numbers   

Description

Input

Output

Sample Input

95.123 12
0.4321 20
5.1234 15
6.7592  9
98.999 10
1.0100 12

Sample Output

548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

Hint

C++

while(cin>>s>>n)

{

...

}

c

while(scanf("%s%d",s,&n)==2) //to  see if the scanf read in as many items as you want

/*while(scanf(%s%d",s,&n)!=EOF) //this also work    */

{

...

}

Source

ACcode：

#include <iostream>
#include <cstring>
#include <cstdio>
#define maxn 999999
using namespace std;
char a[10];
int add[10],ans[maxn],res[maxn],n,doc,len;
void put1(){
    if(doc==-1){
        bool flag=false;
        for(int i=ans[0];i>=1;--i){
            if(ans[i]!=0)flag=true;
            if(flag)printf("%d",ans[i]);
        }
    }
    else {
        doc=5-doc;
        doc*=n;
        int flag1,flag2;
        for(int i=1;i<=ans[0];++i)
            if(ans[i]!=0){
                flag1=i;
                break;
            }
        for(int i=ans[0];i>=1;--i)
            if(ans[i]!=0){
                flag2=i;
                break;
            }
        if(flag2<doc)flag2=doc;
        if(flag1>doc)flag1=doc+1;
        bool flag=false;
        while(flag2>=flag1){
            if(flag2==doc)putchar('.');
            printf("%d",ans[flag2]);
            flag2--;
        }
    }
    putchar('\n');
}
void fun(){
    memset(res,0,sizeof(res));
    for(int i=1;i<=ans[0];i++)  {
        for (int j=1;j<=len;j++)  {
            res[i+j-1]+=ans[i]*add[j];
            if (res[i+j-1]>9)
            {
                res[i+j] += res[i+j-1] / 10;
                res[i+j-1] %= 10;
            }
        }
    }
    if(res[ans[0]+len-1]>9){
        res[ans[0]+len]+=res[ans[0]+len-1]/10;
        res[ans[0]+len-1]%=10;
    }
    ans[0]=ans[0]+len;
    for(int i=1;i<=ans[0];i++)ans[i]=res[i];
}
int main(){
    while(cin>>a>>n){
        if(n==0){
            printf("1\n");
            continue;
        }
        memset(ans,0,sizeof(ans));
        memset(add,0,sizeof(a));
        doc=-1;len=6;
        int i,j,cnt=0;
        for(i=0;i<6;++i){
            if(a[i]=='.'){
                doc=i;
                len=5;
            }
            if(a[i]=='0')cnt++;
        }
        if(cnt==len){
            printf("0\n");
            continue;
        }
        for(i=5,j=1;i>=0;--i)
            if(i!=doc){
                add[j]=(a[i]-'0');
                ans[j]=add[j];
                j++;
            }
            ans[0]=len;
        for(i=1;i<n;++i)
            fun();
            put1();
    }
    return 0;
}

POJ 1001 Exponentiation 求高精度幂

标签：c++   iostream   poj   编程   numbers   

原文地址：http://blog.csdn.net/zp___waj/article/details/47702205