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2001 C 新练习题 答案

2001 C 新练习题 答案

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1 C 1AA2AC3AD4AC 5AB "+ 6A 7A (A main A ) 8A 1A/* A2A */ 9A(1) scanf( ) (2) printf( ) A 3 ] 1AA2AC3AB4AA5AB 6AC7AC8AA9AD10AB 11AB 12AA 13AB 14ACD 15AD 16AC 17AD 18AA 19AC 20AC 21AA 22AA 23AD 24AD 25AB 26AB 27AD 28AA 29AD 30AA 31AA 32AB 33AC 34AA 35AA 36AD 37AA 38AB 39AC 40AC 41AD 42AD 43AC X+ 44A(1) 1 (2) 2 45A(1) 4 (2) 8 46A -16 47A -32768A+32767 48A (1) float (2) double 49A (1) X + (2) X + (3) A 50A 1 51A 26 52A (1) 12 (2) 4 53A (1) 6 (2) 4 (3) 2 54A -60 55A 2 56A (1) 10 (2) 6 57A 5.5 58A 3.5 59A 1 A 60A 1 61A 0 62A 9 63A (1) X + (2) X + (3) A 64A F 65A((((((5*X+3)*X-4)*X+2)*X+1)*X-6)*X+1)*X+10 66A 8.0 67A 13.7 68A int A(A ) 69A double A(A ) 70A (m/10%10)*100 + (m/100)*10 + m%10 A 4 1AD2AD3A[1] B [2] C4AD5AC 6AD7AA8AD9AA A C 10A[1] B [2] B 11AD 12AD 13AB 14AD 15AB 16AD 17AA 18AD 19AA 20AA 21AB 22AB 23AB 24AA 25AD "+ 26A i: dec=A4, oct=177774, hex=fffc, unsigned=65532 27A *3.140000, 3.142*
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28A c: dec=120, oct=170, hex=78, ASCII=x 29A*d(1)=A2*d(2)=  2*d(3)=A2  * *d(4)=177776*d(5)=  177776*d(6)=177776  * 30A*d(1)=A2*d(2)=  2*d(3)=A2  * d(4)=fffe*d(5)=   fffe*d(6)=fffe   * 31A*d(1)=3.50000e+00*d(2)=3.500e+00*d(3)=  3.500e+00* *d(4)=A3.50000e+00*d(5)=A3.50000e+00*d(6)=A3.5000e+00  * 32Ax=1 y=2 *sum*=3 10 Squared is: 100 33A(1)10 (2)     10 (3)56.100000 (4)       3.141600 (5)5.68100e+02 (6)    3.14160e+00 (7)3.1416 (8)       3.1416 34A(1)123.456001 (2)        123.456 (3)123.4560 (4)8765.456700 (5)       8765.457
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