1.题目表述
| Time Limit: 500MS | Memory Limit: 10000K | |
| Total Submissions: 135893 | Accepted: 33256 |
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 */
{
...
}
2.题目大意
给两个数n,R，求n^R.
3.解题方法
高精度乘法或者java的大整数类
4.代码：
1 #include <iostream> 2 #include <cmath> 3 #include <algorithm> 4 #include <cstring> 5 #include <cstdio> 6 #define debug 0 7 using namespace std; 8 void chen(char a[],char b[]) 9 { 10 int i,j,k,l,sum,c[410]={0}; 11 l=strlen(a)+strlen(b); 12 for(i=strlen(b)-1;i>=0;i--) 13 for(j=strlen(a)-1,k=i+j+1;j>=0;j--,k--){ 14 sum=(b[i]-‘0‘)*(a[j]-‘0‘)+c[k]; 15 c[k]=sum%10;c[k-1]+=sum/10; 16 } 17 for(i=c[0]?0:1,j=0;i<l;i++) 18 a[j++]=(c[i]+‘0‘); 19 a[j]=0; 20 } 21 int main() 22 { 23 int r,n,i,j,point,len,d; 24 char s[100],s1[100],c[1000000]; 25 while (scanf("%s%d",&s,&n)!=EOF){ 26 point=0; 27 len=strlen(s); 28 memset(s1,0,sizeof(s1)); 29 i=0; 30 while (s[i]==‘0‘&&i<len){//开始没有预处理这个，遇到0000 1 挂了 31 i++; 32 } 33 d=0; 34 while (i<=len){ 35 s[d]=s[i]; 36 i++; 37 d++; 38 } 39 len=strlen(s); 40 while (s[point]!=‘.‘&&point<len){ 41 point++; 42 } 43 i=point; 44 while (i<len){ 45 s[i]=s[i+1]; 46 i++; 47 } 48 point=len-point-1; 49 point=point*n; 50 if (n!=0){ 51 strcpy(c,s); 52 for (i=2;i<=n;i++){ 53 chen(c,s); 54 } 55 i=point; 56 len=strlen(c); 57 d=0; 58 while (s[d]==‘0‘&&d<len){ 59 d++; 60 } 61 if (d>=len){printf("0");}else 62 if (i>len-d){ 63 printf("."); 64 for (j=1;j<=i-(len-d);j++){ 65 printf("0"); 66 } 67 len=len-1; 68 while (c[len]==‘0‘){ 69 len--; 70 } 71 for (i=d;i<=len;i++){ 72 printf("%c",c[i]); 73 } 74 }else if (i==len-d){ 75 printf("."); 76 len=len-1; 77 while (c[len]==‘0‘){ 78 len--; 79 } 80 for (i=d;i<=len;i++){ 81 printf("%c",c[i]); 82 } 83 }else { 84 i=len-i; 85 len=len-1; 86 while (c[len]==‘0‘){ 87 len--; 88 } 89 for (j=d;j<i+d;j++){ 90 printf("%c",c[j]); 91 } 92 93 if (len>=j){//这里考虑错了0010.00 1 竟然输出“.”，后来改过后就好了. 94 printf("."); 95 for (j=i+d;j<=len;j++){ 96 printf("%c",c[j]); 97 } 98 } 99 } 100 printf("\n"); 101 }else { 102 printf("1\n"); 103 } 104 } 105 return 0; 106 }
原文：http://www.cnblogs.com/nwpuzzh/p/3976178.html