PAT A1024题解——高精度大数相加模板

#include<stdio.h>
#include<string.h>
#include<algorithm>
typedef long long ll;
using namespace std;
struct bign{
    int d[1000];
    int len;
    bign(){
        memset(d,0,sizeof(d));
        len = 0;
    }
};

bign change(char str[]){     //将整数转换为bign
    bign a;
    a.len = strlen(str);
    for(int i=0;i<a.len;i++){
        a.d[i] = str[a.len-1-i] - ‘0‘;
    }
    return a;
}

bool judge(bign a){         //判断是否是回文
    for(int i=0; i <= a.len/2; i++){
        if(a.d[i] != a.d[a.len-1-i]){
            return false;
        }
    }
    return true;
}

bign add(bign a,bign b){               //高精度a+b
    bign c;
    int carry = 0;
    for(int i=0; i < a.len || i < b.len; i++){
        int temp = a.d[i] + b.d[i] + carry;
        c.d[c.len++] = temp % 10;
        carry = temp / 10;
    }
    if(carry != 0){
        c.d[c.len++] = carry;
    }
    return c;
}

void print(bign a){            //输出bign
    for(int i = a.len-1; i >= 0; i--){
        printf("%d",a.d[i]);
    }
    printf("\n");
}

int main(){
    char str[1000];
    int k,num=0;
    while(scanf("%s %d",str,&k)!=EOF){
        bign a = change(str);
        while(num<k && judge(a)==false){
            bign b = a;
            reverse(b.d,b.d+b.len);  //将字符串倒置
            a = add(a,b);
            num++;
        }
        print(a);
        printf("%d\n",num);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/mxj961116/p/10347064.html