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width=515> <BR><BR><BR><BR>两个有300位的正整数相乘<BR><BR>300位的正整数是无法用一般方法存储，计算的。以下用一个程序实现两个有300位的正整数相乘。<BR><BR><BR>#define
N 300
<BR>#include<string.h><BR>#include<stdio.h><BR>main()<BR>{<BR>char
a[N],b[N];<BR>int
an,bn,i,r,ia[N],ib[N],c[2*N],n,m;<BR>scanf("%s%s",a,b);/*输入两个300位以下的正整数*/<BR>an=strlen(a);/*获得第一个正整数的位数*/<BR>bn=strlen(b);/*获得第二个正整数的位数*/<BR>for(i=0;i<an;i++)<BR>ia[an-i-1]=a[i]-'0';/*使a[0]为第一个正整数的个位，a[1]为十位......*/<BR>for(i=0;i<bn;i++)<BR>ib[bn-i-1]=b[i]-'0';<BR>for(i=0;i<2*N;i++)<BR>c[i]=0;<BR>for(i=0;i<an;i++){<BR>for(r=0;r<bn;r++){<BR>c[i+r]=c[i+r]+ia[i]*ib[r];}}/*关键的一步，和我们笔算乘法是一样的，只是其进位没有处理。如果是89*89，那么c[0]=81,c[1]=144,c[2]=64。即有81个1，144个10，64个100，其和为7921。结果正确。*/<BR>for(i=0;i<2*N-1;i++)/*实现进位，从个位开始*/<BR>{<BR>m=c[i];<BR>c[i]=m%10;<BR>c[i+1]=m/10+c[i+1];<BR>}<BR>i=2*N-1;<BR>while(c[i]==0){i--;}/*获得计算结果的位数*/<BR>for(;i>=0;i--)/*输出结果*/<BR>printf("%d",c[i]);<BR>}<BR><BR><BR>--------------------------------------------------------------------------------<BR><BR><BR>输入：
7 8<BR><BR>输出： 56<BR><BR>输入：10
12<BR><BR>输出：120<BR><BR>输入：112233445566778899
998877665544332211<BR><BR>输出：112107482103740987777903741240815689<BR><BR>
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