uva 424 Integer Inquiry (大数加法)
One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers. ``This supercomputer is great,'' remarked Chip. ``I only wish Timothy were here to see these results.'' (Chip moved to a new apartment,once one became available on the third floor of the Lemon Sky apartments on Third Street.) InputThe input will consist of at most 100 lines of text,each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length,and will only contain digits (no VeryLongInteger will be negative). The final input line will contain a single zero on a line by itself. OutputYour program should output the sum of the VeryLongIntegers given in the input. Sample Input123456789012345678901234567890 123456789012345678901234567890 123456789012345678901234567890 0 Sample Output370370367037037036703703703670 题解----就是用数组模拟每位相加,最后再进位,注意数组不要从0,1,2,3开始存, 进位可能需要前面的空间 #include<iostream> #include<cstdio> #include<cstring> using namespace std; int a[105],b[105]; char str[105]; void atoi(char c[])//用数组存每一位 { int p,i; for(i=strlen(c)-1,p=104;i>=0;i--,p--) { b[p]=c[i]-'0'; } } int main() { int i,q=0; memset(a,sizeof(a)); while(cin>>str) { memset(b,sizeof(b)); if(strcmp(str,"0")==0) break; atoi(str); for(i=104;i>=0;i--) { a[i]=b[i]+a[i]; } } for(i=104;i>=0;i--) { a[i-1]=a[i-1]+a[i]/10;//进位 a[i]=a[i]%10;//余数 } for(i=0;i<105;i++) { if(a[i]==0&&q==0) continue; else { cout<<a[i]; q=1; } } puts(""); return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |