Luogu1601 A+B Problem (高精度加法)

蒟蒻复习了下高精

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)

#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)

#define Max(a,b) ((a) > (b) ? (a) : (b))

#define Min(a,b) ((a) < (b) ? (a) : (b))

#define Fill(a,b) memset(a, b, sizeof(a))

#define Swap(a,b) a^=b^=a^=b

#define ll long long

#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")

#define D_e(x)  cout << #x << " = " << x << endl

#define Pause() system("pause")

#else

#define D_e_Line ;

#endif

struct ios{

    template<typename ATP>ios& operator >> (ATP &x){

        x = 0; int f = 1; char c;

        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;

        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();

        x*= f;

        return *this;

    }

}io;

using namespace std;

const int N = 1007;

char strA[N],strB[N];

int a[N],b[N],c[N];

inline void BigAdd(char *strA, char *strB){

	int lenA = strlen(strA + 1), lenB = strlen(strB + 1);

	R(i,1,lenA) a[lenA - i + 1] = strA[i]^'0';

	R(i,1,lenB) b[lenB - i + 1] = strB[i]^'0';

	int len = Max(lenA, lenB);

	R(i,1,len){

		c[i] += a[i] + b[i];

		while(c[i] >= 10){

			++c[i+1];

			c[i] -= 10;

		}

	}

	++len;

	while(c[len] == 0 && len != 1) --len;

	nR(i,len,1)

		printf("%d", c[i]);

}

int main(){

	scanf("%s%s", strA + 1, strB + 1);

	BigAdd(strA, strB);

	return 0;

}

Luogu1601 A+B Problem (高精度加法)的相关教程结束。