#include <iostream>
#include <string>
using namespace std;// 实现大数相加  结果存放在num中 void bigIntergerAdd(string &num, string add) {int goBit = 0; // 存放进位// 先交换下顺序  加数的位数要比较少if (num.length() < add.length()) {string tmp = num;num = add;add = tmp;} string tmp (num.length() - add.length(), '0');add = tmp + add; // 利用string的+号特性  不采用逆序相加法 int len1 = num.length(), len2 = add.length();for (int i = len1 -1 ; i>= 0; --i) {int tmp =  ((num[i] - '0') + (add[i] - '0') + goBit) ;num[i] = tmp% 10 + '0';goBit = tmp/10;}// 特殊情况处理if (goBit != 0) num.insert(0, string(1, (char)goBit +'0'));
} int main(int argc, char** argv) {string s1;string result;int i =0;while (cin>> s1) {if (s1 == "0") {cout<< result<< endl;break;} if (i ==0) {i=1;result = s1;} else bigIntergerAdd(result, s1);} return 0;
}

#include <iostream>
#include <string>
using namespace std; // 大数相乘 string bigIntegerPlus(string res, string plusN) {string ret; if (res.length()< plusN.length()) {string tmp = res;res = plusN;plusN = tmp; }int len1 = res.length(), len2 = plusN.length();for (int i = len2-1; i>=0; --i ) {string tmp(len1, '0'); // 存放相乘的中间结果 int goBit =0;for (int j= len1-1; j >=0; --j) {int mid = (res[j] -'0') * (plusN[i] -'0') + goBit;tmp[j] = mid%10 + '0';goBit = mid /10;}if (goBit != 0) tmp.insert(0, string(1,goBit +'0'));for (int m=0; m< len2 -1-i; ++m)tmp.push_back('0'); // 补位  // 相乘后就相加  大数相加if (i == len2-1)ret = tmp;else {int goBit2 =0; string s(tmp.length() - ret.length() ,'0');ret = s + ret;for (int m = tmp.length()-1; m>=0; --m) {int mid = (tmp[m] -'0')+(ret[m] - '0')  + goBit2;ret[m] = mid %10 +'0';goBit2 = mid/ 10;    }if (goBit2 != 0) ret.insert(0, string(1,goBit +'0'));}     } // 去掉前导0while (ret.length() >1 && ret[0] == '0')ret.erase(0,1);return ret;
}int main(int argc, char** argv) {string res, plusN;while (cin>> res>> plusN) {cout<< bigIntegerPlus(res, plusN)<< endl;}return 0;
}

#include <iostream>
#include <cstring>
using namespace std;void bigDivision(char *src, int num, char sign) {long long rem = 0; // 存放新余数 long long prem = 0; // 原余数char  res[10000] ="";bool flag = true;int k = 0;for (int i=0; i< strlen(src); ++i) {rem = prem * 10/*向后退一位*/ + src[i] - '0';if (rem / num >0 || rem ==0) {res[k++] = rem/ num + '0';prem = rem %num;flag = false;}  else {prem = rem;if (!flag)res[k++] = '0';}}if (sign == '%') {cout<< prem<< endl;return;}for (int i =0; i< k; ++i) cout<< res[i];cout<< endl;
} int main(int argc, char** argv) {char src[10000] = "";int num;char sign;while (scanf("%s %c %d", src, &sign, &num) != EOF) {bigDivision(src, num, sign);}return 0;
}

#include <iostream>
#include <string>
using namespace std;// 求幂  思路: 先变成整数相乘   然后根据小数的位数 结合幂  算出小数点该结果字符串的位置 即可string bigIntegerPlus(string src, string num) {string tmp = src;for (int i =num.length() -1; i >= 0 ; --i) {string mid(tmp.length(),'0');int goBit =0;for (int j =  tmp.length()-1; j >= 0; --j) {int tm = goBit + (tmp[j] -'0')* (num[i] - '0');mid[j] = tm% 10 +'0';goBit = tm  /10;}for (int q = num.length()-1; q> i; --q) mid.push_back('0');if (goBit != 0) mid.insert(0, string(1, (char)goBit +'0'));// 加法运算if (i == num.length()-1)src = mid;else {goBit =0;string s(mid.length() - src.length(), '0');src = s + src;for (int j = mid.length()-1; j>=0; --j) {int tm = (mid[j] - '0') +(src[j] - '0') + goBit;src[j] = tm %10 + '0';goBit = tm /10;}if (goBit !=0) src.insert(0, string(1, (char)goBit +'0'));}    }return src;
}int main(int argc, char** argv) {string str;while ( getline(cin, str)) {// 分割出待积数 和 幂  以及小数点位置int i =0;int index = 0;// 小数位置int count = 0;//幂次数string num;while ( i< str.length()) {if ( str[i] != ' ') {if (str[i] == '.')index = i;elsenum.push_back(str[i]);++i;continue;}while ( !isdigit(str[i])) ++i;if (i + 1 == str.length())count = str[i] - '0';else count = (str[i] - '0') * 10 + str[i+1] - '0';        break;        }index = num.length() - index;string res = num;for (int i =0; i< count-1; ++i) {res = bigIntegerPlus( res, num);}index = index * count;res.insert(res.length() - index, string(".")); while (res.length() >1 && res[0] == '0')res.erase(0, 1);for (int i =res.length()-1; i>=0; --i) {if (res[i] == '0' )res.erase(i, i+1);else break;}cout<< res<< endl;    }return 0;
}