B1016 部分A B

一、算法笔记的方法

#include<cstdio>
int main(){
    long long a,b,da,db;
    scanf("%lld%lld%lld%lld",&a,&da,&b,&db);
    long long pa = 0,pb = 0;
    while(a != 0)
    {
        if(a%10 == da) pa = pa * 10 + da;
        a = a / 10;
    }
    while(b != 0)
    {
        if(b%10 == db) pb = pb * 10 + db;
        b = b / 10;
    }
    printf("%lld\n",pa+pb);
    return 0;
}

二、我的方法

#include<cstdio>
#include<string>
using namespace std;
int main(){
    long long A,B;
    int DA,DB;
    int countA = 0,countB =0;
    int resultA = 0;
    int resultB = 0;
    scanf("%lld %d %lld %d",&A,&DA,&B,&DB);
    //计算各自D的数量
    while(A!=0){
        int l = A%10;
        A = A/10;
        if (l == DA){
            countA++;
        }
    }
    while(B!=0){
        int l = B%10;
        B = B/10;
        if (l == DB){
            countB++;
        }
    }
    for(int i = 0;i<countA;i++){
        resultA = resultA + DA;
        DA = DA*10;
    }
    for(int i = 0;i<countB;i++){
        resultB = resultB + DB;
        DB = DB*10;
    }
    printf("%d",resultA + resultB);
    return 0;
}