http://ac.jobdu.com/problem.php?pid=1016
来源:互联网 发布:查士丁尼瘟疫 知乎 编辑:程序博客网 时间:2024/05/01 20:23
- 题目描述:
- 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……
- 输入:
- 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。
- 输出:
- 对每个测试用例输出1行,即火星表示法的A+B的值。
- 样例输入:
1,0 2,14,2,0 1,2,01 10,6,4,2,10 0
- 样例输出:
1,0,11,1,1,01,0,0,0,0,0
- #include<iostream>
- #include<cstdio>
- #include<memory.h>
- #include<algorithm>
- #include<cstring>
- #include<cmath>
- #include<cstdlib>
- using namespace std;
- int prime[70];
- int a[70];
- int b[70];
- int c[70];
- bool isPrime(int n){
- int q = sqrt(n);
- for(int i=2;i<=q;++i){
- if(n%i==0){
- return false;
- }
- }
- return true;
- }
- int main(){
- //freopen("in.txt", "r", stdin);
- int k=1;
- for(int i=2;1;i++){
- if(isPrime(i)){
- prime[k++] = i;
- }
- if(k==68)
- break;
- }
- string s1, s2;
- while(true){
- memset(a, 0, sizeof(a));
- memset(b, 0, sizeof(b));
- memset(c, 0, sizeof(c));
- cin>>s1>>s2;
- int len1 = s1.length();
- int k = 1;
- int t;
- bool iszero = true;
- for(int i=len1-1;i>=0;i-=2){
- t = s1[i]-'0';
- while(i-1>=0 && s1[i-1]!=',')
- t = (s1[--i]-'0')*10+t;
- a[k++] = t;
- if(a[k-1]!=0)
- iszero = false;
- }
- if(iszero)
- break;
- iszero = true;
- int len2 = s2.length();
- k = 1;
- for(int i=len2-1;i>=0;i-=2){
- t = s2[i]-'0';
- while(i-1>=0 && s2[i-1]!=',')
- t = (s2[--i]-'0')*10+t;
- b[k++] = t;
- if(b[k-1]!=0)
- iszero = false;
- }
- if(iszero)
- break;
- int len = max((len1-1)/2+1, (len2-1)/2+1);
- for(int i=1;i<=len+1;++i){
- c[i] += a[i]+b[i];
- if(c[i]>=prime[i]){
- c[i+1] = c[i]/prime[i];
- c[i] = c[i]%prime[i];
- }
- }
- bool flag = true;
- for(int i =len+1;i>1;i--){
- if(flag){
- if(c[i]==0)
- continue;
- else
- flag = false;
- }
- printf("%d,", c[i]);
- }
- printf("%d\n", c[1]);
- }
- //fclose(stdin);
- return 0;
- }
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