POJ 1001Exponentiation解题报告——求高精度幂

http://poj.org/problem?id=1001

这个道题比我想象中要难很多，做了很久，修改了很多次才搞定。

算法要点：

1. 使用字符串模拟数据相乘法.multiply函数就是将a*b的内容，放到字符串s中

2. 很重要：乘法都是从低位到高位乘，使用数组，需要把一个数（使用字符串表示）进行倒序，才方便计算。可以在最后都计算完以后，再把结果再倒序回来。

3. 很重要：要考虑边界的测试，需要准备一组测试数据。我在下面的代码里面增加了测试函数。如果#define TEST，那么会走测试流程，测试在data.txt中的数据。一定要保证所有的这些数据测试通过，才能在POJ上提交，否则往往会有问题。

data.txt我放在文章的最后。

/*
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.
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 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
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX 512
//#define TEST
char s[MAX],b[MAX],a[6];
int n,loc;

//1. 找到小数点位置，根据幂次计算小数点的位置
//2. 将s中小数点的位置替换为""
//3. 如果有小数点，将s中最后为0的部分干掉
void init(){
int ln;
int i;
char *p = strstr(s,".");

if(NULL!=p){
ln = strlen(s) - (p-s)
cca2
- 1;

//remove the ending '0'
for(i=strlen(s)-1;i>=ln;i--){
if('0'==s[i])
ln--;
else
break;
}
//remove '.'
for(i=0;i<ln;i++){
p[i]=p[i+1];
}
p[i]=0;
}else{
ln = 0;
}

//去掉开头的0
for(i=0;i<strlen(s);i++){
if('0'!=s[i]){
break;
}
}
if(i==strlen(s)){//all zero
s[0]=0;
}else{
int j=0;
for(i=i;i<=strlen(s);i++){
s[j++]=s[i];
}
}

loc = ln*n;
}
//s = a*b
void multiply(){
int flag=0;
for(int i=0;i<strlen(a);i++)
for(int j=0;j<strlen(b);j++){
s[i+j] += (a[i]-'0')*(b[j]-'0');
s[i+j+1] += s[i+j]/10;
s[i+j] %=10;
}

for(int i=(MAX-1);i>=0;i--){
if(flag){
s[i] +='0';
}else{
if(0!=s[i-1])
flag=1;
}
}
}
void reverse(char *s){
int size=strlen(s);

char ch;
for(int i=0;i<size/2;i++){
ch=s[i];
s[i]=s[size-i-1];
s[size-i-1]=ch;
}
}
void calculate(){
reverse(s);
strcpy(a,s);
while(--n){
strcpy(b,s);
memset(s,0,MAX);
multiply();
}
reverse(s);
}
void printToB(){
int i;
char *p=b;
//需要考虑s的长度还不如小数点的长度
if(loc>strlen(s)){
int zero=loc-strlen(s);
sprintf(p,".");
p +=strlen(p);
for(i=0;i<zero;i++){
sprintf(p,"0");
p +=strlen(p);
}
for(i=0;i<strlen(s);i++){
sprintf(p,"%c",s[i]);
p +=strlen(p);
}
}else{
for(i=0;i<strlen(s);i++){
if(loc>0 && (loc==(strlen(s)-i))) {
sprintf(p,".");
p +=strlen(p);
}
sprintf(p,"%c",s[i]);
p +=strlen(p);
}
}
}

#if defined(TEST)
int main(){
int i,j;
char problem[512];
char answer[512];
char *file="data.txt";
FILE *fh=fopen(file,"r");
if(NULL==fh){
printf("Cannot open %s\n",file);
return 1;
}
while(!feof(fh)){
fgets(problem,512,fh);
fgets(answer,512,fh);
for(i=0;i<strlen(problem);i++){
if(' '!=problem[i])
s[i]=problem[i];
else{
s[i]=0;
break;
}
}
for(i=i;i<strlen(problem);i++)
if(' '!=problem[i])
break;
j=0;
for(i=i;i<strlen(problem);i++)
if(' '!=problem[i])
a[j++]=problem[i];
else{
a[j++]=0;
break;
}
n = atoi(a);

init();
//special case
if(0==strlen(s)){
printf("%d\n",0);
continue;
}else if(0==n){
printf("%d\n",1);
continue;
}
calculate();

printToB();
//check
if('\n'==answer[strlen(answer)-1])
answer[strlen(answer)-1]=0;
if(strlen(b)!=strlen(answer)){
printf("%s wrong!!!!\n",problem);
printf("expected: %s\n",answer);
printf("computed: %s\n",b);
return 1;
}else{
for(i=0;i<strlen(b);i++){
if(b[i]!=answer[i]){
printf("%s wrong!!!!\n",problem);
printf("expected: %s\n",answer);
printf("computed: %s\n",b);
printf("wrong location: %d\n",i);
return 1;
}

}
}
}
}
#else
int main(){
while(EOF!=(scanf("%s %d",s,&n))){
init();
//special case
if(0==strlen(s)){
printf("%d\n",0);
continue;
}else if(0==n){
printf("%d\n",1);
continue;
}

calculate();
printToB();
printf("%s",b);
printf("\n");
}

return 0;
}
#endif

data.txt:

00.000 20

0

5.1004 15

41120989454.314570363993506408035342551967503175087477761156936917581824

6.0092 20

3769929003093657.32196296074639653207879581595068906610421189179087477999341087617646994418302976

90.909 20

1486406551798253233999195346707087657544.775949857477432123445196642299879937148848818861186416630801

1.0001 20

1.00200190114048465507876775325986797847727972597775238761550448451140019000200001

54.120 20

46468190221655199272477781205783832.2745385034134800709627347812689616306176

000.10 20

.00000000000000000001

12.010 20

3898164373852177448724.9596914878392975482722144801842193624001

1.1000 20

6.72749994932560009201

00.100 20

.00000000000000000001

.10000 25

.0000000000000000000000001

.98765 25

.73295397043235459290597253263360363155207373749758041835495018812263204038846757925736111183672158735201714336872100830078125

0000.1 25

.0000000000000000000000001

0.0001 1

.0001

0.0000 20

0

11.001 20

673974233702250167359.962694826025671047351081040835313734121365641162362990220001

110000 20

67274999493256000920100000000000000000000000000000000000000000000000000000000000000000000000000000000

.00001 25

.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

.00010 23

.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

1.0000 25

1

2.0000 25

33554432

6.0000 25

28430288029929701376

99.999 25

99975002999770012649468717709519310815545705768715.426520024799744573673126042964184298069822900531298735002299997000002499999

1 0

1

0 1

0

.00 1

0

95.123 12

548815620517731830194541.899025343415715973535967221869852721

0.4321 20

.00000005148554641076956121994511276767154838481760200726351203835429763013462401

5.1234 15

43992025569.928573701266488041146654993318703707511666295476720493953024

6.7592 9

29448126.764121021618164430206909037173276672

98.999 10

90429072743629540498.107596019456651774561044010001

1.0100 12

1.126825030131969720661201

.00001 1

.00001

.12345 1

.12345

0001.1 1

1.1

1.1000 1

1.1

10.000 1

10

000.10 1

.1

000000 1

0

000.00 1

0

.00000 0

1

000010 1

10

000.10 1

.1

0000.1 1

.1

00.111 1

.111

0.0001 1

.0001

0.0001 3

.000000000001

0.0010 1

.001

0.0010 3

.000000001

0.0100 1

.01

0.0100 3

.000001

0.1000 1

.1

0.1000 3

.001

1.0000 1

1

1.0000 3

1

1.0001 1

1.0001

1.0001 3

1.000300030001

1.0010 1

1.001

1.0010 3

1.003003001

1.0100 1

1.01

1.0100 3

1.030301

1.1000 1

1.1

1.1000 3

1.331

10.000 1

10

10.000 3

1000

10.001 1

10.001

10.001 3

1000.300030001

10.010 1

10.01

10.010 3

1003.003001

10.100 1

10.1

10.100 3

1030.301

99.000 1

99

99.000 3

970299

99.001 1

99.001

99.001 3

970328.403297001

99.010 1

99.01

99.010 3

970593.059701

99.100 1

99.1

99.100 3

973242.271

99.998 1

99.998

99.998 3

999940.001199992