| Time Limit: 500MS | Memory Limit: 10000K | |
| Total Submissions: 127046 | Accepted: 31014 |
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
与UVa 748是同一道题(http://www.cnblogs.com/lzj-0218/p/3528164.html)，但是在输出的地方有一点小小的差别，就是这POJ的这道题中，如果算出来是个整数的话，不要输出小数点
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 5 using namespace std; 6 7 int main() 8 { 9 int r; 10 int a[200],b[10],c[200]; 11 char s[20]; 12 13 14 while(gets(s)) 15 { 16 memset(a,0,sizeof(a)); 17 memset(b,0,sizeof(b)); 18 a[0]=1; 19 20 int t=0,pos; 21 for(int i=5;i>=0;i--) 22 if(s[i]!=‘.‘) 23 b[t++]=s[i]-‘0‘; 24 else 25 pos=i; 26 sscanf(&s[7],"%d",&r); 27 28 for(int k=1;k<=r;k++) 29 { 30 memset(c,0,sizeof(c)); 31 for(int i=0;i<t;i++) 32 { 33 int u=0; 34 for(int j=0;i+j<200;j++) 35 { 36 int temp=c[j+i]; 37 c[j+i]=(a[j]*b[i]+c[j+i]+u)%10; 38 u=(a[j]*b[i]+temp+u)/10; 39 } 40 } 41 for(int i=0;i<200;i++) 42 a[i]=c[i]; 43 } 44 45 int Start,End; 46 for(Start=199;a[Start]==0;Start--); 47 for(End=0;a[End]==0;End++); 48 if(Start<(5-pos)*r) 49 { 50 putchar(‘.‘); 51 Start=(5-pos)*r-1; 52 } 53 if(End>(5-pos)*r) 54 End=(5-pos)*r; 55 for(int i=Start;i>=End;i--) 56 if(i==(5-pos)*r&&i!=End) 57 printf("%d.",a[i]); 58 else 59 printf("%d",a[i]); 60 putchar(‘\n‘); 61 } 62 63 return 0; 64 }
原文：http://www.cnblogs.com/lzj-0218/p/3528169.html
