两个多项式相乘时间复杂度的怎么算，请给出
#include#include#include#include/*显示帮助信息*/voidhelp(){printf("\ \n");printf(" \n");exit(0);}/*显示错误信息*/voidprinterror(errno,num,base1)charerrno,*num,*base1;{switch(errno){case1:printf("\ (%s)isvalid!!!\n",num,base1);break;case2:printf("\ (%s)isinvalid!!!\n%s\n",base1," >=2andradix36))printerror(2,""...全部
#include#include#include#include/*显示帮助信息*/voidhelp(){printf("\ \n");printf(" \n");exit(0);}/*显示错误信息*/voidprinterror(errno,num,base1)charerrno,*num,*base1;{switch(errno){case1:printf("\ (%s)isvalid!!!\n",num,base1);break;case2:printf("\ (%s)isinvalid!!!\n%s\n",base1," >=2andradix36))printerror(2,"",base1);/*有效吗?*/ibase2=atoi(base2);/*数基2*/if((ibase236))printerror(2,"",base2);/*有效吗?*/for(j=0;j=48)&&(num[j]10){/*求每个字母所代表的十进制值*/if((num[j]>=''A'')&&(num[j]=''a'')&&(num[j]<=m))k=num[j]-''a''+10;elseprinterror(1,num,base1);}elseprinterror(1,num,base1);inum+=k*(int)r;/*累加计算结果*/}/*输出转换结果*/printf("%s(%d)=%s(%d)\n",num,ibase1,ltoa(inum,temp,ibase2),ibase2);}/*主程序*/main(argc,argv)intargc;char*argv[];{staticcharnum[10],base1[10],base2[10];printf("(TNS)TranslatorofNumberSystem1。
0Copyright(c)1995DongZhanshan\n");switch(argc){case1:case2:case3:help();break;case4:strcpy(num,argv[1]);strcpy(base1,argv[2]);strcpy(base2,argv[3]);transnum(num,base1,base2);}。
收起
