Problem Description
读入两个不超过25位的火星正整数A和B,计算A+B。需要注意的是:在火星上,整数不是单一进制的,第n位的进制就是第n个素数。例如:地球上的10进制数2,在火星上记为“1,0”,因为火星个位数是2进制的;地球上的10进制数38,在火星上记为“1,1,1,0”,因为火星个位数是2进制的,十位数是3进制的,百位数是5进制的,千位数是7进制的……
Input
测试输入包含若干测试用例,每个测试用例占一行,包含两个火星正整数A和B,火星整数的相邻两位数用逗号分隔,A和B之间有一个空格间隔。当A或B为0时输入结束,相应的结果不要输出。
Output
对每个测试用例输出1行,即火星表示法的A+B的值。
Sample Input
1,0 2,1 4,2,0 1,2,0 1 10,6,4,2,1 0 0
Sample Output
1,0,1 1,1,1,0 1,0,0,0,0,0
将两种算法结合起来求解
#include<iostream>
#include<cstring>
#include<cstdio>
#include<stack>
#include<cmath>
#include<algorithm>
#include<ctype.h>
using namespace std;
int prime[26];
char num1[1000];
char num2[1000];
void is_prime()
{
int cnt=0;
for(int i=2;; i++)
{
int ok=1;
int m=int(sqrt(i)+0.5);
for(int j=2; j<=m; j++)
if(i%j==0)
{
ok=0;
break;
}
if(ok)prime[cnt++]=i;
if(cnt>=26)break;
}
}
int main()
{
is_prime();
//freopen("datain.txt","r",stdin);
while(~scanf("%s%s",num1,num2)){
if(!strcmp(num1,"0")&&!strcmp(num2,"0"))break;
int pos1=strlen(num1)-1;
int pos2=strlen(num2)-1;
stack<int>s;
int r=0,digit=0;
while(pos1>=0||pos2>=0){
int weight=0,x=0;
for(;pos1>=0;pos1--){
if(isdigit(num1[pos1]))x+=(num1[pos1]-'0')*pow(10,weight++);
else break;
}
weight=0;
for(;pos2>=0;pos2--){
if(isdigit(num2[pos2]))x+=(num2[pos2]-'0')*pow(10,weight++);
else break;
}
s.push((x+r)%prime[digit]);
r=(x+r)/prime[digit];
pos1--;pos2--;digit++;
}
if(r) s.push(r);
printf("%d",s.top());s.pop();
while(!s.empty()) printf(",%d",s.top()),s.pop();
printf("\n");
}
return 0;
}
顺便帖一下刘老师的大数算法
#include<cstdio>
#include<cstring>
#include<vector>
#include<iostream>
#include <algorithm>
#include<string>
using namespace std;
struct BigInteger {
static const int BASE = 100000000;
static const int WIDTH = 8;
vector<int> s;
BigInteger(long long num = 0) { *this = num; } // 构造函数 给构造函数的参数提供默认值
BigInteger operator = (long long num) { // 赋值运算符
s.clear();
do {
s.push_back(num % BASE);
num /= BASE;
} while (num > 0);
return *this;
}
BigInteger operator = (const string& str) { // 赋值运算符
s.clear();
int x, len = (str.length() - 1) / WIDTH + 1;
for (int i = 0; i < len; i++) {
int end = str.length() - i * WIDTH;
int start = max(0, end - WIDTH);
sscanf(str.substr(start, end - start).c_str(), "%d", &x);
s.push_back(x);
}
return *this;
}
BigInteger operator + (const BigInteger& b) const {
BigInteger c;
c.s.clear();
for (int i = 0, g = 0; ; i++) {
if (g == 0 && i >= s.size() && i >= b.s.size()) break;
int x = g;
if (i < s.size()) x += s[i];
if (i < b.s.size()) x += b.s[i];
c.s.push_back(x % BASE);
g = x / BASE;
}
return c;
}
};
ostream& operator << (ostream &out, const BigInteger& x) {
out << x.s.back();
for (int i = x.s.size() - 2; i >= 0; i--) {
char buf[20];
sprintf(buf, "%08d", x.s[i]);
for (int j = 0; j < strlen(buf); j++) out << buf[j];
}
return out;
}
istream& operator >> (istream &in, BigInteger& x) {
string s;
if (!(in >> s)) return in;
x = s;
return in;
}
#include<set>
#include<map>
set<BigInteger> s;
map<BigInteger, int> m;
int main() {
BigInteger y;
BigInteger x = y;
BigInteger z = 123;
BigInteger a, b;
cin >> a >> b;
cout << a + b << "\n";
//cout << BigInteger::BASE << "\n";
return 0;
}