学步园

简单模拟题，主要就是“地址”，“地址的地址”，“地址里的数据”这几个概念别搞乱脑子就好。C/C++老手应该闭着眼睛也能写出正确的程序来：）

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//////////////////////////////////////////////////////////////////////////<br /> // 10033 - Interpreter<br /> // Copyright (c) 2010 by Bo-wen Feng<br /> // balonfan@gmail.com<br /> //////////////////////////////////////////////////////////////////////////</p> <p>#include <set><br /> #include <map><br /> #include <list><br /> #include <queue><br /> #include <stack><br /> #include <vector><br /> #include <string><br /> #include <sstream></p> <p>#include <cstdio><br /> #include <cmath><br /> #include <limits><br /> #include <utility></p> <p>#ifndef ONLINE_JUDGE<br /> #include <fstream><br /> std::ifstream cin("in.txt");<br /> std::ofstream cout("out.txt");<br /> // std::ofstream cout(stdout);<br /> #else<br /> #include <iostream><br /> #endif</p> <p>typedef long long llong;<br /> typedef unsigned long long ullong;<br /> typedef long double ldouble;</p> <p>using namespace std;</p> <p>#define MAX_MEM 1001<br /> #define MAX_REG 11</p> <p>char memx[MAX_MEM][4];<br /> int regx[MAX_REG];</p> <p>int getMem(int i)<br /> {<br /> return (memx[i][0] * 10 + memx[i][1]) * 10 + memx[i][2];<br /> }</p> <p>void setMem(int i, int value)<br /> {<br /> memx[i][0] = value / 100;<br /> value %= 100;<br /> memx[i][1] = value / 10;<br /> memx[i][2] = value % 10;<br /> }</p> <p>void Run()<br /> {<br /> char skip[1024];<br /> int ip = 0;</p> <p> memset(memx, 0, sizeof(memx));<br /> memset(regx, 0, sizeof(regx));</p> <p> for(; !cin.eof(); ++ip)<br /> {<br /> cin>> memx[ip];<br /> memx[ip][0] -= '0';<br /> memx[ip][1] -= '0';<br /> memx[ip][2] -= '0';</p> <p> cin.getline(skip, 1024);</p> <p> char next = cin.peek();<br /> if(next < '0' || next > '9')<br /> break;<br /> }</p> <p> int maxIP = ip;<br /> int nRunLines = 0;</p> <p> char* ins;</p> <p> for(ip = 0; <img src="https://admin.xuebuyuan.com/wp-includes/images/smilies/icon_wink.gif" alt=";)" class="wp-smiley" /><br /> {<br /> ins = memx[ip];<br /> ++nRunLines;<br /> ++ip;</p> <p> switch(ins[0])<br /> {<br /> case 1:<br /> break;</p> <p> case 2:<br /> regx[ins[1]] = ins[2];<br /> break;</p> <p> case 3:<br /> regx[ins[1]] = (regx[ins[1]] + ins[2]) % 1000;<br /> break;</p> <p> case 4:<br /> regx[ins[1]] = (regx[ins[1]] * ins[2]) % 1000;<br /> break;</p> <p> case 5:<br /> regx[ins[1]] = regx[ins[2]];<br /> break;</p> <p> case 6:<br /> regx[ins[1]] = (regx[ins[1]] + regx[ins[2]]) % 1000;<br /> break;</p> <p> case 7:<br /> regx[ins[1]] = (regx[ins[1]] * regx[ins[2]]) % 1000;<br /> break;</p> <p> case 8:<br /> regx[ins[1]] = getMem(regx[ins[2]]);<br /> break;</p> <p> case 9:<br /> setMem(regx[ins[2]], regx[ins[1]]);<br /> break;</p> <p> case 0:<br /> if(regx[ins[2]] != 0)<br /> ip = regx[ins[1]];<br /> break;<br /> }</p> <p> if(ins[0] == 1)<br /> break;<br /> }</p> <p> cout<< nRunLines<< '/n';<br /> }</p> <p>int main(int argc, char* argv[])<br /> {<br /> int n;</p> <p> cin>> n;</p> <p> for(int i = 0; i < n; ++i)<br /> {<br /> if(i > 0)<br /> cout<< '/n';</p> <p> Run();<br /> }</p> <p> return 0;<br /> }</p> <p>