思路：利用高精度加法求出和然后根据每一位要求的进制进行计算 前四个过了，自己造了一个发现不对

样例：

输入：120062+2150
标准输出：1,0,2,2,0
实际输出：1,2,2,2,0

代码：

#include<iostream>
#include<cstring>
#include<climits>
using namespace std;
string s;
int sy[10] = {0 , 2 , 3 , 5 , 7 , 11 , 13}; // 对应 个 十 百 千 万 十万
int as = INT_MAX , bs = INT_MAX , f;
int a[102034] , b[102034] , c[102034] , d[102034];
string v; string w;
int len;
void jia(){
	len = max(v.length() , w.length());
	for(int i = len; i >= 1; i --){
		c[i] = a[i] + b[i];
	}
	
	int len1 = 1;
	
	for(int i = len1; i <= len; i ++){
		if(c[i] >= 10){
			c[i + 1] += c[i] / 10;
			c[i] %= 10;
		}
	}
	while(c[len + 1] != 0) len ++;
//	for(int i = len; i >= 1; i --) cout << c[i];
}

int main(){
	cin >> s;
	for(int i = 0; i < s.length(); i ++){
		if(s[i] == '+') f = 1;
		if(s[i] >= '0' && s[i] <= '9'){
			if(as == INT_MAX && bs == INT_MAX) as = i;
			else{
				if(f == 0) continue;
				bs = i;
				break;
			}
		}
	}


	for(int i = as; i < bs; i ++)
		if(s[i] >= '0' && s[i] <= '9')
			v += s[i];

	for(int i = bs; i < s.length(); i ++)
		if(s[i] >= '0' && s[i] <= '9')
			w += s[i];

	for(int i = 0; i < v.length(); i ++) a[v.length() - i] = v[i] - '0';
	for(int i = 0; i < w.length(); i ++) b[w.length() - i] = w[i] - '0';

	jia();
//	cout << endl;
	int len2 = 1;
	for(int i = 1 , j = 1; i <= len , j <= len; i ++ , j ++){
		if(i >= 5){
			int temp = c[i + 1] * 10 + c[i];
			if(temp >= sy[j]){
				d[len2] += temp % sy[j]; // 1
				d[len2 + 1] = temp / sy[j]; // 1
				c[i + 1] = d[len2 + 1]; // c[i + 1] == c[i];
				len2 ++;
			}
		}
		else{
			d[len2] += c[i] % sy[j];
			d[len2 + 1] = c[i] / sy[j];
			len2 ++;
		}
	}

	for(int i = 1 , j = 1; i <= len2 , j <= len; i ++ , j ++){
		if(i >= 5){
			int temp = d[i + 1] * 10 + d[i];
			if (temp >= sy[j]){
				i ++;
				d[i + 1] += temp / sy[j];
				d[i] = temp % sy[j];
			}
		}
		if(i < 5 && d[i] >= sy[j]){
			d[i + 1] += d[i] / sy[j];
			d[i] %= sy[j];
		}
//		cout << d[i] << " " << sy[j] << endl;
	}
	
	
	while(d[len2] == 0 && len2 > 1) len2 --;

	for(int i = len2; i >= 1; i --){
		if(i != len2){
			cout << ",";
		}
		cout << d[i];
	}
	return 0;
}

感谢各位大佬！！