北大OJ 1001题 Exponentiation

作者：林子木

大体思路就是

用递归的方法求幂的次数

其中乘法，使用大数乘法的原理，不拘泥本题的六位的输入 

其中整型字符串 和 字符串转整型 采用自己写的函数，如果使用系统自带的，速率会跟高些。

AC的时间是16MS 内存是232K

这里的主要问题是有块内存不能释放在，在注释的

//free(s2);

//free(s1);

这里的内存指向是最后的char *Int2Str(int *n,int len)中产生的内存

char *MyPow(char *s,int n){char *s1,*s2,*res;if(n == 1){return Mutli(s,"1");}else if(n == 2){return Mutli(s,s);}else if(n > 2){s1 = MyPow(s,n>>1);if(n%2 != 0){s2 = MyPow(s,n-(n>>1));res = Mutli(s1,s2);//free(s2);}else{res = Mutli(s1,s1);}//free(s1);return res;}}

代码如下：

#include<stdio.h>#include <stdlib.h>#include<string.h>#include <math.h>#define LENGTH 6char *MyPow(char *s,int n);char *Mutli(char *s1,char *s2);int *Str2Int(char *s,int lenArr);char *Int2Str(int *n,int len);int main(){char realNumString[LENGTH]; int n;int piontLocation;//标记小数位后有几位char *numPointer;while(scanf("%s %d",realNumString,&n) != EOF){int len,lenRes;int i,j;char tmp[LENGTH];char *res;piontLocation = 0;//处理后面的无用的零 len = strlen(realNumString); numPointer = realNumString +  len - 1; while(*numPointer == '0') { *numPointer = '\0'; numPointer--; } //处理前面无用的零 numPointer = realNumString + strspn(realNumString, "0");//只有使用指针才能如此操作 len = strlen(numPointer); j = 0; for( i=0; i<len; i++) {if( *(numPointer + i) == '.')piontLocation = len - i - 1;elsetmp[j++] = *(numPointer + i); } tmp[j] = '\0';//末尾加上 strcpy(realNumString,tmp+strspn(tmp, "0"));//再次去除前面没用的零 res = MyPow(realNumString,n); len = strlen(realNumString); //如果是整数 if (piontLocation == 0) { printf("%s\n",res); } //如果只有小数没整数 else if (len <= piontLocation) { lenRes = strlen(res); i = piontLocation*n - lenRes; if (i == 0) {printf(".%s\n",res); } else {printf(".");while (i--){printf("0");}printf("%s\n",res); } } //其他 else { lenRes = strlen(res);for( i=0; i<lenRes; i++){if(i == lenRes-piontLocation*n)printf(".");printf("%c",res[i]);}printf("\n"); }}}char *MyPow(char *s,int n){char *s1,*s2,*res;if(n == 1){return Mutli(s,"1");}else if(n == 2){return Mutli(s,s);}else if(n > 2){s1 = MyPow(s,n>>1);if(n%2 != 0){s2 = MyPow(s,n-(n>>1));res = Mutli(s1,s2);//free(s2);}else{res = Mutli(s1,s1);}//free(s1);return res;}}char *Mutli(char *s1,char *s2){int len1,len2;int lenArr1,lenArr2,lenTotal;int *n1,*n2,*nRes;char *res;int i,j;int carry;len1 = strlen(s1);len2 = strlen(s2);res = (char *)malloc((len1 + len2 + 1)*sizeof(char)); //判断s2是否为1if (strncmp(s2, "1", 1) == 0 && len2 == 1){strcpy(res,s1);return res;}lenArr1 = (len1%4==0) ? (len1/4) : (len1/4+1);lenArr2 = (len2%4==0) ? (len2/4) : (len2/4+1);lenTotal = lenArr1 + lenArr2;nRes= (int *)malloc(lenTotal*sizeof(int));memset(nRes,0,sizeof(int)*(lenTotal));//初始化n1 = Str2Int(s1,lenArr1);n2 = Str2Int(s2,lenArr2);//计算相乘for( i=lenArr1-1; i>=0; i--){for( j=lenArr2-1; j>=0;j--){nRes[i+j+1] += n1[i] * n2[j] ; //注意要+1 本位nRes[i+j] += nRes[i+j+1] / 10000;nRes[i+j+1] %= 10000;}}free(n1);free(n2);free (res);//现在res指的内存是这此次mutli创出来的，下面的res指向的是另外的地址res = Int2Str(nRes,lenTotal);free(nRes);return res;}int *Str2Int(char *s,int lenArr){int *res;int i,j;int lenStr;res = (int *)malloc(lenArr*sizeof(int));//分配空间memset(res,0,lenArr*sizeof(int));//初始化j = lenArr;lenStr = strlen(s);for( i=lenStr-1; i>=0; i--){if((lenStr-i-1)%4 != 0) //lenStr-i 从最后开始数{res[j] +=  (int)(s[i]-'0') * pow(10.0,(lenStr-i-1)%4);//lenStr-i-1 第一位是n的0次幂}else{res[--j] += s[i]-'0';}}return res;}char *Int2Str(int *n,int len){//这里我是使用10000进制每个数组单位里有四个字符int i,j;char *res;res = (char *)malloc((len*4+1)*sizeof(char));memset(res,'0',len*4*sizeof(char));res[len*4] = '\0';//最后加上'\0'for( i=0; i<len; i++)for( j=0; j<4; j++){n[i] %= (int)pow(10.0,4-j);res[i*4+j] = n[i]/pow(10.0,3-j) + '0';}res = res + strspn(res,"0");return res;}