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<TD style="FONT-SIZE: 12px" vAlign=top
width=515> <BR><BR><BR><BR>兩個有300位的正整數相乘<BR><BR>300位的正整數是無法用一般方法存儲，計算的。以下用一個程序實現兩個有300位的正整數相乘。<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);/*輸入兩個300位以下的正整數*/<BR>an=strlen(a);/*獲得第一個正整數的位數*/<BR>bn=strlen(b);/*獲得第二個正整數的位數*/<BR>for(i=0;i<an;i++)<BR>ia[an-i-1]=a[i]-'0';/*使a[0]為第一個正整數的個位，a[1]為十位......*/<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];}}/*關鍵的一步，和我們筆算乘法是一樣的，只是其進位沒有處理。如果是89*89，那么c[0]=81,c[1]=144,c[2]=64。即有81個1，144個10，64個100，其和為7921。結果正確。*/<BR>for(i=0;i<2*N-1;i++)/*實現進位，從個位開始*/<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--;}/*獲得計算結果的位數*/<BR>for(;i>=0;i--)/*輸出結果*/<BR>printf("%d",c[i]);<BR>}<BR><BR><BR>--------------------------------------------------------------------------------<BR><BR><BR>輸入：
7 8<BR><BR>輸出： 56<BR><BR>輸入：10
12<BR><BR>輸出：120<BR><BR>輸入：112233445566778899
998877665544332211<BR><BR>輸出：112107482103740987777903741240815689<BR><BR>
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