OJ_1119
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#include <iostream>#include <string>using namespace std;void func(){ string s; char sum[101]; for(int i=0;i<101;i++) sum[i]='0'; while(cin>>s) { if(s=="0") { int i=100; while(sum[i]=='0')i--; for(;i>=0;i--) { cout<<sum[i]; } cout<<endl; break; } int j=0; int inc=0; int i=s.size()-1; for(;i>=0;i--) { int newk=sum[j]-'0'+s[i]-'0'+inc; inc=newk/10; sum[j]=newk%10+'0'; j++; } while(inc||sum[j]!='0') { int newk=sum[j]-'0'+inc; inc=newk/10; sum[j]=newk%10+'0'; j++; } }}int main(int argc, char *argv[]){ //printf("Hello, world\n");func();//system("pause");return 0;}
大数加法
- 题目描述:
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.)
- 输入:
The 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.
- 输出:
Your program should output the sum of the VeryLongIntegers given in the input.
- 样例输入:
1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678900
- 样例输出:
370370367037037036703703703670
- 提示:
注意输入数据中,VeryLongInteger 可能有前导0