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poj1001(高精度乘法)
1.题目表述
Exponentiation
| Time Limit: 500MS | Memory Limit: 10000K | |
| Total Submissions: 135893 | Accepted: 33256 |
Description
Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.
This problem requires that you write a program to compute the exact value of Rn where R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.
This problem requires that you write a program to compute the exact value of Rn where R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.
Input
The input will consist of a set of pairs of values for R and n. The R value will occupy columns 1 through 6, and the n value will be in columns 8 and 9.
Output
The output will consist of one line for each line of input giving the exact value of R^n. Leading zeros should be suppressed in the output. Insignificant trailing zeros must not be printed. Don‘t print the decimal point if the result is an integer.
Sample Input
95.123 120.4321 205.1234 156.7592 998.999 101.0100 12
Sample Output
548815620517731830194541.899025343415715973535967221869852721.0000000514855464107695612199451127676715483848176020072635120383542976301346240143992025569.92857370126648804114665499331870370751166629547672049395302429448126.76412102161816443020690903717327667290429072743629540498.1075960194566517745610440100011.126825030131969720661201
Hint
If you don‘t know how to determine wheather encounted the end of input: s is a string and n is an integer
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 }
poj1001(高精度乘法)
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