Emag eht htiw Em Pleh【模拟】

与这道题刚好相反

他要做的是把原来输出的，变成现在输入的，原来输入的，变成现在输出的。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define eps 1e-8
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(x, y) make_pair(x, y)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
map<char, int> mp;
char s[17][36]=
{{"+---+---+---+---+---+---+---+---+\n\0"},  //0{"|...|:::|...|:::|...|:::|...|:::|\n\0"},  //1{"+---+---+---+---+---+---+---+---+\n\0"},  //2{"|:::|...|:::|...|:::|...|:::|...|\n\0"},  //3{"+---+---+---+---+---+---+---+---+\n\0"},  //4{"|...|:::|...|:::|...|:::|...|:::|\n\0"},  //5{"+---+---+---+---+---+---+---+---+\n\0"},  //6{"|:::|...|:::|...|:::|...|:::|...|\n\0"},  //7{"+---+---+---+---+---+---+---+---+\n\0"},  //8{"|...|:::|...|:::|...|:::|...|:::|\n\0"},  //9{"+---+---+---+---+---+---+---+---+\n\0"},  //10{"|:::|...|:::|...|:::|...|:::|...|\n\0"},  //11{"+---+---+---+---+---+---+---+---+\n\0"},  //12{"|...|:::|...|:::|...|:::|...|:::|\n\0"},  //13{"+---+---+---+---+---+---+---+---+\n\0"},  //14{"|:::|...|:::|...|:::|...|:::|...|\n\0"},  //15{"+---+---+---+---+---+---+---+---+\n\0"},  //16
};
char w[205], b[205];
void init()
{mp['K'] = mp['k'] = 1;mp['Q'] = mp['q'] = 2;mp['R'] = mp['r'] = 3;mp['B'] = mp['b'] = 4;mp['N'] = mp['n'] = 5;mp['P'] = mp['p'] = 6;
}
inline int fid_x(int x) { return (x - 1) * 4 + 2; }
inline int fid_y(int y) { return (8 - y) * 2 + 1; }
int main()
{init();int len, x, y;scanf("White: ");scanf("%s\n", w);len = (int)strlen(w);for(int i=0, j; i<len; ){j = i;while(w[j] ^ ',') j++;if(j - i == 3){x = w[i + 1] - 'a' + 1;y = w[i + 2] - '0';s[fid_y(y)][fid_x(x)] = w[i];}else{x = w[i] - 'a' + 1;y = w[i + 1] - '0';s[fid_y(y)][fid_x(x)] = 'P';}i = j + 1;}scanf("Black: ");scanf("%s\n", b);len = (int)strlen(b);for(int i=0, j; i<len; ){j = i;while(b[j] ^ ',') j++;if(j - i == 3){x = b[i + 1] - 'a' + 1;y = b[i + 2] - '0';s[fid_y(y)][fid_x(x)] = b[i] + 32;}else{x = b[i] - 'a' + 1;y = b[i + 1] - '0';s[fid_y(y)][fid_x(x)] = 'p';}i = j + 1;}for(int i=0; i<17; i++) printf("%s", s[i]);return 0;
}

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Emag eht htiw Em Pleh【模拟】

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