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C语言练习册_ç­”æ&

C语言练习册_ç­”æ&

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ˆ5E 1¢ ¢ C ; ƒþ í A¢ case B¢ -ac C¢ _53 D¢ ab-c 2 k # D A¢ -0. \55’ C¢ 0x2a3 103’ a> | b> | (‘4’-‘0’) – a*b …&k @»ƒ ¢ ((int)’4’-(int)’0’)-(double)a*(double)b_¢ x=2¢ y=3¢ x=y==3> | ______ 1 _________¢ a j 12> | a+=a -=a*a ¢ a ¢ A¢ 144 B¢ -264 C¢ 264 D¢ 0 x¢ y¢ z> 1> | w=++x || ++y &&++z ¢ x¢ y¢ z> | A¢ x=1¢ y=1¢ z=2 B¢ x=2¢ y=2¢ z=2 C¢ x=1¢ y=2¢ z=1 D¢ x=2¢ y=1¢ z=1 7> | C C fb# 1 a=7+b+c=a+7 ; a=7+ b++ =a+7 ; a=7+b , b++ , a+7 a=7+b , c=a+7 ; C # ( # # k & A¢ % B¢ / C¢ < D¢ ! 9> b a b a x - + ) (sin 2 ¢ C> | __ sin(x)*sin(x)*(a+b)/(a-b) 10¢ C kk A j B j C [ D…& 11¢ x y z ≥ ≥ ° C (x >=y) && (y >=z) (x >=y) AND (y >=z) (x >= y >= z) (x >=y) & (y >=z) 12> if j a ¢ 0…& | a < >0 B¢ !a C¢ a=0 D¢ a C… 1999 ¢ 2 ¢ 1
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13| 10 < x < 100 ¢ x < 0 ¢ C | _ (x>10&&x<100)||x<0 _ ¢ 14¢ a¢ b¢ c ¢ int ¢ a=3¢ b=4¢ c=5 E 0 ¢ a’ && ‘b’ B¢ a<=b a || b + c && b-c ! ((a<b) && !c ||1 ) 15 [ int i=65536 ; printf (“%d \n”, i ) ; 65536 B¢ 0 C…& D¢ -1 16¢ ch ¢ char | A ch = (ch >= ‘A’ && ch <= ‘Z’) ? (ch + 32) : ch ¢ A a Z z 17¢ x ¢ y ¢ int ¢ x=100¢ y=200 ¢ printf (“%d”, (x , y)) ; ¢ 200 100 100 200 D …& 18¢ x ¢ int…& for ( x=3 ; x<6 ; x++) printf ((x%2) ? (“ * * %d ”) : (“ # # %d \n”) , x ) ; * * 3 # # 3 # # 4 * * 4 * * 5 # # 5 # # 3 * * 3 # # 4 * * 4 # # 5 * * 5 19¢ i ¢ s …& | for(i=0 , s=0 ; i+s<10 , i<10 ; i++ , s++) ; printf (“%d ,%d\n” , i , s) ; ¢ 11 , 11 5 , 5 6 , 6 10 , 10 20¢ main () { int x=3 , y=0 , z=0 ; if (x=y+z) printf (“* * * * “) ; else printf (“# # # # “) ; } k 6 A `ÇI¹ * * * * C… 1999 ¢ 2 ¢ 2
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CHw |Y… &… & # # # # 21¢ main () { int x=3 ; do { printf (“%d\n”, x - = 2) ; } while ( ! ( - - x) ) ; } k 1 1 ¢ 2 3 ¢ 0 D 22¢ main () { int x=023 ; printf ( “%d\n”, - - x) ; } k A¢ 17 B¢ 18 C¢ 23 D¢ 24 23¢ main () { int x=10 , y=10 ; printf (“%d %d\n”, x - - , - - y ) ; 21 } | A¢ 10 10 B¢ 9 9 C¢ 9 10 D¢ 10 9 24¢ main () { int x ; scanf ( “%d”, &x) ; if ( x++ > 5) printf ( “%d\n”, x ) ; else printf ( “%d\n”, x - - ) ; } ‘eg 5 …& A¢ 7 B¢ 6 C¢ 5 D¢ 4 25¢ main () { int a ; printf (“%d\n”, (a=3*5 , a*4 , a+5 ) ) ; } ‘eg a A¢ 65 B¢ 20 C¢ 15 D¢ 10 26¢ main () { int a=-1 , b=4 , k ; k=(a++ <= 0) && ( ! (b - - <= 0) ) ; printf ( “%d %d %d\n”, k , a , b ) ; } , 0 0 3 0 1 2 1 0 3 1 1 2 27Hw | áU²áQ+ Q 6 ¢ 4 | C… 1999 ¢ 2 ¢ 3
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main () { int x ; scanf (“%d”, &x ) ; if ( x++ > 5) printf (“%d “, x ) ; else printf (“%d\n”, x - - ) ; } A¢ 7 ¢ 5 B¢ 6 ¢ 3 C¢ 7 ¢ 4 D¢ 6 ¢ 4 28# (# # k + = < - = ) 0 ( 1 ) 0 ( 0 ) 0 ( 1 x x x y = - < = ) 0 ( 0 ) 0 ( 1 ) 0 ( 1 x x x y = - < = ) 0 ( 1 ) 0 ( 1 ) 0 ( 0 x x x y = < - = ) 0 ( 0 ) 0 ( 1 ) 0 ( 1 x x x y y = -1 ; if ( x != 0) if ( x > 0 ) y=1 ; else y=0 ; 29¢ #include <stdio.h> main() { int a ; float b , c ; scanf (“%2d %3f %4f “, &a , &b, &c ) ; printf( “\n a=%d , b=%f , c=%f \n “ , a , b , c) ; } …& 9876543210 <CR>pT | a=98 , b=765 , c=4321 a=10 , b=432 , c=8765 a=98 , b=765.000000 , c=4321.000000 a=98 , b=765.0 , c=4321.0 30¢ printf( “%d \n “, strlen (“ ATS\n012\1\\ “ )) ; s 11 10 9 8 31¢ #include <stdio.h> main () { int a , b ; for ( a=1 , b=1 ; a<=100 ; a++) { if ( b>=20 ) break ; if ( b%3 == 1 ) { b += 3 ; continue ; } C… 1999 ¢ 2 ¢ 4
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b - = 5 ; } printf ( “%d \n “, a ) ; } !
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