让CPU占有率曲线听你指挥

解法一：简单的解法

int main() { for (;;) { for (int i = 0; i < 9600000; i++) ; Sleep(10); } return 0; }

解法二：使用GetTickCount()和Sleep()

include <iostream> include <windows.h> using namespace std; const DWORD busyTime = 10; //10ms const DWORD idleTime = busyTime; //same ratio will lead to 50% cpu usage DWORD startTime = 0; int main() { while (TRUE) { DWORD startTime = GetTickCount(); //busy loop while ((GetTickCount() - startTime) <= busyTime); //idle loop Sleep(idleTime); } return 0; }

解法三：能动态适应的解法

//C# code static void MakeUsage(float level) { PerformanceCounter p = new PerformanceCounter("Processor", "Processor Time", "_Total"); while (TRUE) { if (p.NextValue() > level) { System.Threading.Thread.Sleep(10); } } }

解法四：正弦曲线

//C++ code to make task manager generate sine graph #include "windows.h" #include "stdlib.h" #include "math.h" //把一条正弦曲线0~2∏之间的弧度等分成200份进行抽样，计算每个抽样点的振幅 //然后每隔300ms的时间取下一个抽样点，并让CPU工作对应振幅的时间 const int SAMPLINE_COUNT = 200; //抽样点数量 const double PI = 3.1415926535; //pi值 const int TOTAL_AMPLITUDE = 300; //每个抽样点对应的时间片 int main() { DWORD busySpan[SAMPLINE_COUNT]; int amplitude = TOTAL_AMPLITUDE / 2; double radian = 0.0; double radianIncrement = 2.0 / (double)SAMPLINE_COUNT; //抽样弧度的增量 for (int i = 0; i < SAMPLINE_COUNT; i++) { busySpan[i] = (DWORD)(amplitude + (sin(PI * radian) * amplitude)); radian += radianIncrement; //printf("%d\t%d\n", busySpan[i], TOTAL_AMPLITUDE - busySpan[i]); } DWORD startTime = 0; for (int j = 0; ; j = (j + 1) % SAMPLINE_COUNT) { startTime = GetTickCount(); while ((GetTickCount() - startTime) <= busySpan[j]) ; Sleep(TOTAL_AMPLITUDE - busySpan[j]); } return 0; }