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luckymark
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Mar 5, 2017
| #include<stdio.h> | ||
| #include<windows.h> | ||
| #define SPACE 50 | ||
| void show(char *pos); |
P02_isprime.c
Outdated
| mark = 1; //若被整除，标记mark为1 | ||
| break; | ||
| } | ||
| }if (mark == 1) |
| for(int i=2;i<=1000;i++) | ||
| { | ||
| if (isprinme(i)) | ||
| printf("%5d", i); |
p05_allPrimes.c
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| #include<stdio.h> | ||
| #include<stdlib.h> | ||
| #include<time.h> | ||
| int isprinme(int a); |
| } | ||
| if (mark == 1) //检查标记 | ||
| return 0; | ||
| else return 1; |
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这里改为？表达式更清真一点（这是我们项目组内部的说法，意思是代码漂亮干净）
| if (a == prime[i] + prime[j]) //遍历两个100以内的素数组合之和，与a作比较 | ||
| { | ||
| printf("%d满足哥德巴赫猜想\n",a); | ||
| goto b; //跳出遍历循环 |
p06_Goldbach.c
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| mark = 1; //若被整除，标记mark为1 | ||
| break; | ||
| } | ||
| }if (mark == 1) |
| ret_val = fgets(st, n, stdin); | ||
| if (ret_val) | ||
| { | ||
| while (st[i] != '\n'&&st[i] != '\0') |
| @@ -0,0 +1,528 @@ | |||
| #include<stdio.h> | |||
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