编程之美随书源代码-中文注释.doc

1、1第 1 章游 戏 之 乐游戏中碰到的题目1. 代码清单 1-1int main()for(; ; for(int i = 0; i level) System.Threading.Thread.Sleep(10); 代码清单 1-4/ C+ code to make task manager generate sine graph#include “Windows.h“#include “stdlib.h“#include “math.h“const double SPLIT = 0.01;const int COUNT = 200;const double PI = 3.14159265;

2、const int INTERVAL = 300;int _tmain(int argc, _TCHAR* argv)DWORD busySpanCOUNT; / array of busy times DWORD idleSpanCOUNT; / array of idle timesint half = INTERVAL / 2;double radian = 0.0;for(int i = 0; i HALF_BITS_LENGTH)/ 这个数字表示右 bits的mask，在二进制表示中，它表示00001111。#define RSET(b, n) (b = (LMASK for(LSE

3、T(b, 1); LGET(b) 0);m_nCakeCnt = nCakeCnt;/ 初始化烙 饼数组m_CakeArray = new intm_nCakeCnt; Assert(m_CakeArray != NULL);for(int i = 0; i m_nMaxSwap)return;/ 如果已 经排好序，即翻转完成，输出结果if(IsSorted(m_ReverseCakeArray, m_nCakeCnt)if(step pCakeArrayi)return false;return true;/ 翻转烙饼 信息/ void Revert(int nBegin, int nEnd

4、)Assert(nEnd nBegin);int i, j, t;8/ 翻 转烙饼信息for(i = nBegin, j = nEnd; i = 0; j-)for(int i = 0; i optij)optij = x;9return optV0;代码清单 1-10intV + 1T + 1 opt; / 子问题的记录项表，假设从i到T 种饮料中，/ 找出容量总和为V 的一个方案，满意度最多能够达到/ opt（V，i，T-1），存储于optVi，/ 初始化时opt中存储值为-1 ，表示该子问题尚未求解。int Cal(int V, int type)if(type = T)if(V = 0

5、)return 0;elsereturn -INF;if(V ret)ret = temp;return optVtype = ret;代码清单 1-11int nPerson; / nPersoni表示到第i 层的乘客数目int nFloor, nMinFloor, nTargetFloor;nTargetFloor = -1;for(i = 1; i nFloor) 10nMinFloor = nFloor;nTargetFloor = i;return(nTargetFloor, nMinFloor);代码清单 1-12int nPerson; / nPersoni表示到第i层的乘客数目

6、int nMinFloor, nTargetFloor;int N1, N2, N3;nTargetFloor = 1;nMinFloor = 0;for(N1 = 0, N2 = nPerson1, N3 = 0, i = 2; i = N; i+)N3 += nPersoni;nMinFloor += nPersoni * (i - 1);for(i = 2; i = N; i+)if(N1 + N2 N3)nTargetFloor = i;nMinFloor += (N1 + N2 - N3);N1 += N2;N2 = nPersoni;N3 -= nPersoni;elsebreak

7、;return(nTargetFloor, nMinFloor);代码清单 1-13int nMaxColors = 0, i, k, j;for(i = 0; i N; i+)for(k = 0; k nMaxColors; k+)isForbiddenk = false;for(j = 0; j i; j+)if(Overlap(bj, ej, bi, ei)isForbiddencolorj = true;for(k = 0; k nMaxColors; k+)if(!isForbiddenk)break;if(knMaxColors)colori = k;elsecolori = nMaxColors+; return nMaxColors；