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http://acm.pku.edu.cn/JudgeOnline/problem?id=1001
求高精度冪的題目，題目的難度在于處理如同 .00010 的輸入
和輸出的前導(dǎo)后導(dǎo)零的問(wèn)題，還有小數(shù)點(diǎn)的位置可能在處理輸入時(shí)候被影響 寫(xiě)了好久代碼，但是同學(xué)用JAVA只需要一會(huì)就能搞定了，郁悶... 附上AC代碼：
1 Source Code
2
3Problem: 1001 User: hongtaozhy
4Memory: 304K Time: 0MS
5Language: G++ Result: Accepted
6
7Source Code
8#include<stdio.h>
9#include<string.h>
10#include<math.h>
11void reverse(char*str);
12void multi(char*num1,char*num2,char*result);
13 char a[100];
14 char str[100];
15 char sum[200];
16 char tem[200];
17 char tem2[100];
18  int main() {
19 int flag;
20 int n;
21 int key;
22 int flag2;
23 int kk;
24//freopen("a.txt","r",stdin);
25//freopen("a2.txt","w",stdout);
26  while(scanf("%s%d",a,&n)==2){
27 memset(str,0,sizeof(str));
28 kk=0;
29 flag=-1;
30 flag2=6;
31 int g=0;
32 int j=0;
33 int t = 0;
34 if(a[0]=='.'){ g=1;}
35 for(int i = 0 ; i < 6 ; i++){
36 if(a[i]=='0'&&kk==0)continue;
37 kk=1;
38 a[t++]=a[i];
39 }
40 a[t]='\0';
41
42 if(a[0]=='.') {
43 for(int i =strlen(a);i>=0;i--)
44 a[i+1]=a[i];
45 a[0]='0';
46 }{
47 int i;
48 for(i=0 ; i < strlen(a);i++ )
49 if(a[i]=='.') break;
50 if(i==strlen(a)) {a[i]='.';a[i+1]='0';a[i+2]='\0';}}
51
52 for(int i=0;i<strlen(a);i++)
53 if(a[i]=='.') { flag=i;kk=1;}
54 else {
55 if(a[i]=='0'&&flag!=-1&&flag2==strlen(a)) flag2=i;
56 if(a[i]!='0'&&flag!=-1) flag2=strlen(a);
57 str[j++]=a[i];
58 }
59
60 int ne=strlen(a)-flag-1;
61 for(j = strlen(str)-1 ; j >= 0 ; j-- ){
62 if(str[j] == '0'&&ne--){
63 str[j]='\0';
64 }
65 else break;
66 }
67 flag=flag2-flag-1;
68 //接入結(jié)束
69 if(g==1)
70 printf(".");
71 memset(sum,0,sizeof(sum));
72 memset(tem,0,sizeof(tem));
73 memset(tem2,0,sizeof(tem2));
74 key=0;
75 sum[0]='1';
76 sum[1]='\0';
77 for(int i=0; i < n ; i++)
78 {
79 if(key!=1){
80 key=1;
81 strcpy(tem2,str);
82 multi(sum,tem2,tem);
83 memset(sum,0,sizeof(sum));
84 memset(tem2,0,sizeof(tem2));
85 }
86 else{
87 strcpy(tem2,str);
88 key=0;
89 multi(tem,tem2,sum);
90 memset(tem,0,sizeof(tem));
91 memset(tem2,0,sizeof(tem2));
92 }
93 }
94 if(key==1){
95 for(int i = 0 ; tem[i] != '\0' ; i++ ){
96 if( tem[i] == '0' && i == 0 ) continue;
97 if(i==strlen(tem)-(flag*n)&&g!=1) printf(".");
98 printf("%c",tem[i]);
99 }
100 }
101 else{
102 for(int i = 0 ; sum[i] != '\0' ; i++ ){
103 if( sum[i] == '0' && i == 0 ) continue;
104 if(i==strlen(sum)-(flag*n)&&g!=1) printf(".");
105 printf("%c",sum[i]);
106 }
107 }
108 printf("\n");
109 }
110return 0;
111 }
112void multi(char*num1,char*num2,char*result)
113{
114 int i,j,len1,len2,len;
115
116 len1=strlen(num1);
117 len2=strlen(num2);
118 reverse(num2);
119 reverse(num1);
120
121 for (i=0;i<len1;i++)
122 num1[i] -= '0';
123 for (i=0;i<len2;i++)
124 num2[i] -= '0';
125
126 for (i=0;i<len2;i++)
127 for (j=0;j<len1;j++)
128 {
129 len=i+j;
130 result[len] += num2[i]*num1[j];
131 result[len+1] += result[len]/10;
132 result[len] %= 10;
133 }
134 len=len1+len2-1;
135 for (i=0;i<len;i++)
136 result[i] += '0';
137 if (result[len]) result[len] += '0';
138 reverse(result);
139}
140
141void reverse(char*str)
142{
143 int i;
144 char c;
145 for (i=0;i<strlen(str)/2;i++)
146 {
147 c=str[i];
148 str[i]=str[strlen(str)-i-1];
149 str[strlen(str)-i-1]=c;
150 }
151}
152
153
154
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