【問(wèn)題描述】設(shè)計(jì)一個(gè)結(jié)構(gòu)體類型，描述進(jìn)程的結(jié)構(gòu)，如圖1所示。然后定義一個(gè)結(jié)構(gòu)體數(shù)組存儲(chǔ)如圖2所示的4個(gè)進(jìn)程。初始時(shí)進(jìn)程的狀態(tài)都是就緒狀態(tài)即pState=1，從優(yōu)先級(jí)最高的進(jìn)程開(kāi)始執(zhí)行，將執(zhí)行的進(jìn)程狀態(tài)pState由就緒狀態(tài)改為執(zhí)行狀態(tài)即pState=2，并輸出當(dāng)前進(jìn)程信息（包括進(jìn)程的每一個(gè)屬性的信息），然后該進(jìn)程pCPU--，直到為0，將該進(jìn)程狀態(tài)改為pState=3，表示進(jìn)程執(zhí)行完畢。輸出當(dāng)前進(jìn)程信息（包括進(jìn)程的每一個(gè)屬性的信息）。接下來(lái)執(zhí)行剩余就緒狀態(tài)進(jìn)程中優(yōu)先級(jí)最高的進(jìn)程，直到?jīng)]有就緒狀態(tài)進(jìn)程為止。

【樣例輸入】沒(méi)有輸入，以上信息在程序中直接賦值。
【樣例輸出】

1 :chrome.exe? 101? 10 2? 2? google chrome

2 :chrome.exe? 101? 10 2? 1? google chrome

3 :chrome.exe? 101? 10 2? 0? google chrome

3 :chrome.exe? 101? 10 3? 0? google chrome

4 :360.exe? ? ?102? 7? 2? 0? 360 safe protect

4 :360.exe? ? ?102? 7? 3? 0? 360 safe protect

5 :QQ.exe? ? ? 103? 5? 2? 3? tencet QQ

6 :QQ.exe? ? ? 103? 5? 2? 2? tencet QQ

7 :QQ.exe? ? ? 103? 5? 2? 1? tencet QQ

8 :QQ.exe? ? ? 103? 5? 2? 0? tencet QQ

8 :QQ.exe? ? ? 103? 5? 3? 0? tencet QQ

9 :CPPIDE.exe? 104? 3? 2? 1? CFree 5.0

10:CPPIDE.exe? 104? 3? 2? 0? CFree 5.0

10:CPPIDE.exe? 104? 3? 3? 0? CFree 5.0

#includeusing namespace std;
struct process
{
	string pname;
	int pid;
	int ppriority;
	int pstate;
	int pcpu;
	string descripition;
};
int main()
{
	int count=0;
	process a[4]={{"chrome.exe",101,10,2,2,"google chrome"},{"360.exe",102,7,2,0,"360 safe protect"},{"QQ.exe",103,5,2,3,"tencet QQ"},{"CPPIDE.exe",104,3,2,1,"CFree 5.0"}};
	for(int i=0;i<4;i++)
	{
		for(int j=a[i].pcpu;j>=0;j--)
		{
			count++;
			cout<
【問(wèn)題描述】設(shè)計(jì)一個(gè)結(jié)構(gòu)體類型，描述進(jìn)程的結(jié)構(gòu)，如圖1所示。然后定義一個(gè)結(jié)構(gòu)體數(shù)組或鏈表存儲(chǔ)如圖2所示的4個(gè)進(jìn)程。初始時(shí)進(jìn)程的狀態(tài)都是就緒狀態(tài)即pState=1，從優(yōu)先級(jí)最高的進(jìn)程開(kāi)始執(zhí)行，將執(zhí)行的進(jìn)程狀態(tài)pState由就緒狀態(tài)改為執(zhí)行狀態(tài)即pState=2。然后不斷循環(huán)，每循環(huán)一次，就緒狀態(tài)的進(jìn)程優(yōu)先級(jí)增加1，其余不變；執(zhí)行狀態(tài)的進(jìn)程優(yōu)先級(jí)減3且其pCPU減1；當(dāng)pCPU為0時(shí)，該進(jìn)程執(zhí)行完畢，其進(jìn)程狀態(tài)修改為停止即pState=3，其余不再變化。直到所有進(jìn)程pCPU都為0，則循環(huán)結(jié)束。
【樣例輸入】沒(méi)有輸入，以上信息在程序中直接賦值。
【樣例輸出】
1?:360.exe?????102??8??1??1??360?safe?protect
1?:QQ.exe??????103??6??1??4??tencet?QQ
1?:CPPIDE.exe??104??4??1??2??CFree?5.0

2?:chrome.exe??101??8??1??2??google?chrome
2?:360.exe?????102??5??2??0??360?safe?protect
2?:QQ.exe??????103??7??1??4??tencet?QQ
2?:CPPIDE.exe??104??5??1??2??CFree?5.0

3?:chrome.exe??101??5??2??1??google?chrome
3?:360.exe?????102??5??3??0??360?safe?protect
3?:QQ.exe??????103??8??1??4??tencet?QQ
3?:CPPIDE.exe??104??6??1??2??CFree?5.0

4?:chrome.exe??101??6??1??1??google?chrome
4?:360.exe?????102??5??3??0??360?safe?protect
4?:QQ.exe??????103??5??2??3??tencet?QQ
4?:CPPIDE.exe??104??7??1??2??CFree?5.0

5?:chrome.exe??101??7??1??1??google?chrome
5?:360.exe?????102??5??3??0??360?safe?protect
5?:QQ.exe??????103??6??1??3??tencet?QQ
5?:CPPIDE.exe??104??4??2??1??CFree?5.0

6?:chrome.exe??101??4??2??0??google?chrome
6?:360.exe?????102??5??3??0??360?safe?protect
6?:QQ.exe??????103??7??1??3??tencet?QQ
6?:CPPIDE.exe??104??5??1??1??CFree?5.0

7?:chrome.exe??101??4??3??0??google?chrome
7?:360.exe?????102??5??3??0??360?safe?protect
7?:QQ.exe??????103??4??2??2??tencet?QQ
7?:CPPIDE.exe??104??6??1??1??CFree?5.0

8?:chrome.exe??101??4??3??0??google?chrome
8?:360.exe?????102??5??3??0??360?safe?protect
8?:QQ.exe??????103??5??1??2??tencet?QQ
8?:CPPIDE.exe??104??3??2??0??CFree?5.0

9?:chrome.exe??101??4??3??0??google?chrome
9?:360.exe?????102??5??3??0??360?safe?protect
9?:QQ.exe??????103??2??2??1??tencet?QQ
9?:CPPIDE.exe??104??3??3??0??CFree?5.0

10:chrome.exe??101??4??3??0??google?chrome
10:360.exe?????102??5??3??0??360?safe?protect
10:QQ.exe??????103??-1?2??0??tencet?QQ
10:CPPIDE.exe??104??3??3??0??CFree?5.0
#includeusing namespace std;
struct process
{
	string pname;
	int pid;
	int ppriority;
	int pstate;
	int pcpu;
	string descripition;
};
int main()
{
	int count=0;
	process a[4]={{"chrome.exe",101,10,1,3,"google chrome"},{"360.exe",102,7,1,1,"360 safe protect"},{"QQ.exe",103,5,1,4,"tencet QQ"},{"CPPIDE.exe",104,3,1,2,"CFree 5.0"}};
	while(1)
	{
		int max=0;
		for(int j=0;j<4;j++)
		{
			
			if(a[0].pcpu==0&&a[1].pcpu==0&&a[2].pcpu==1&&a[3].pcpu==0)
			{
				max=2;
			}
			else
			{
				if(a[j].pcpu!=0)
			 {
				if(a[max].ppriority
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