求2000~3000年之间年月日各个位上数字之和为素数的有多少？ 比如（2012—2—13，和为2+0+1+2+2+1+3=11）
\u4e00\u4e2a\u8d28\u6570\u76843\u500d\u4e0e\u53e6\u4e00\u4e2a\u8d28\u6570\u76842\u500d\u4e4b\u548c\u7b49\u4e8e2000\uff0c\u90a3\u4e48\u8fd9\u4e24\u4e2a\u8d28\u6570\u7684\u548c\u662f\u591a\u5c11\uff1f\u89e3 \u56e0\u4e3a2000\u4e3a\u4e24\u4e2a\u5947\u6570\u6216\u5076\u6570\u7ec4\u6210
\u4e00\u4e2a\u6570\u76842\u500d\u4e0d\u53ef\u80fd\u4e3a\u5947\u6570
\u6240\u4ee5\u4e00\u4e2a\u6570\u76843\u500d\u4e3a\u5076\u6570
\u6240\u4ee5\u7b2c\u4e00\u4e2a\u8d28\u6570\u4e3a2\uff0c\u79ef\u4e3a6
\u6240\u4ee5\u53e6\u4e00\u4e2a\u8d28\u6570\u4e3a\uff082000-6\uff09/2=997
\u6240\u4ee5\u4e24\u4e2a\u6307\u6570\u5206\u522b\u4e3a2\u548c997
\u8bbe\u8fd9\u4e24\u4e2a\u8d28\u6570\u5206\u522b\u4e3a m \u548cn\u3002
3m + 2n = 2000
\u6240\u4ee5
m = 2*(1000-n)/3
n = 1000 - 3m/2
\u4e0d\u770b\u7b2c\u4e00\u4e2a\u5f0f\u5b50\uff0c\u53ea\u770b\u7b2c\u4e8c\u4e2a\u5f0f\u5b50\u3002\u56e0\u4e3a\u8fd9\u4e2a\u5f0f\u5b50\u80fd\u63d0\u4f9b\u5f88\u76f4\u63a5\u91cd\u8981\u7684\u4fe1\u606f\u3002
n = 1000 - 3m/2
\u4e3a\u4fdd\u8bc1n\u662f\u6574\u6570\uff0c\u90a3\u4e48 3m/2 \u5fc5\u987b\u662f\u6574\u6570
\u4e3a\u4fdd\u8bc1 3m/2 \u662f\u6b63\u6570\uff0c\u90a3\u4e48 m\u5fc5\u987b\u662f\u5076\u6570\u3002
\u800cm\u540c\u65f6\u53c8\u662f\u8d28\u6570\u3002
\u56e0\u6b64 m \u53ea\u80fd\u662f2\u3002\u6bd42\u5927\u7684\u5076\u6570\u90fd\u4e0d\u662f\u8d28\u6570\u3002
m=2 \u4ee3\u5165\u7b2c\u4e8c\u4e2a\u5f0f\u5b50
n = 1000 - 3m/2 = 997
\u7ecf\u68c0\u9a8c\uff0c 997 \u4e5f\u662f\u8d28\u6570\u3002
一是判断相等要用==；
二是，你老老实实写y==4||y==6||y==9||y==11吧。
fun3里if语句是否为双等号==
绛旓細浜屾槸锛屼綘鑰佽佸疄瀹炲啓y==4||y==6||y==9||y==11鍚с
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绛旓細<%Dim i, tmpInt, yCounttmpInt = 0: yCount = 0For i = 2000 to 3000 If (i Mod 4 = 0 And i Mod 100 <> 0) Or (i Mod 400 = 0) Then If tmpInt = 10 Then tmpInt = 1: Response.Write("" & CStr(i) & "") Else tmpInt = tmpInt + 1: Response.Write(CStr...
绛旓細int main(){int sum = 0;for (int i = 2000; i <= 3000; i += 4) {if ((sum % 100 == 0) && (sum % 400 > 0))continue;sum++;printf("%4d骞翠娇闂板勾\n", i);}printf("涓鍏辨湁%d骞存槸闂板勾", sum);system("pause");return 0;} ...
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