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The reporter managed to get a(n) ( ) interview with the Prime Minister.
A. inclusive
B. exclusive
C. impressive
D. executive
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A. occasionB. easeC. investigationD. interview
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But for her mother's sudden illness, she would never think of breaking this _________with you.
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阅读以下说明和流程图8-11，完成程序(n)处的语句写在对应栏内。[说明]对于数学上一个猜想：任何自然数平方的36倍等于两对孪生素数的和。初始的情形如下：12×36=(5+7)+(11+13)22×36=(29+31)+(41+43)32×36=(11+13)+(149+151)再往下，N取4，5，6，时，只要N不太大，也都可以找到N(上标)2×36等于两对孪生素数的和。但是当N是一个任意的正整数时，证明N2×36总是等于两对孪生素数的和，这还是一个目前尚未解决的问题。甚至当考察的数较大时，找出一组符合条件的两对孪生素数都是计算量相当大的工作。每尝试一次，都要作4次是否是素数的判断，要作许多次的尝试，才可能找到一组解。下面流程图设计了一种优化算法来对这个猜想进行验证。仔细阅读流程图8-11，完成程序部分。[程序部分]main (){int t, i, j, prime_index; is_p rime:long n, p, p1, p2, p3, p4, s, s1;long primes [ 16000 ];for (n=1; n＜98; ++n){t=0;s= n* n* 36;prime_index= 2;primes[0]=2; primes[1]=3;for (p=5: p＜=s/2; p=p+2){is_p rime= 1;for ( i=1;(1)++i)if ( p%primes [i] = = 0 ) is_p rime= 0;if ( is_p rime){(2)}}for ( i=1; (3)++i){(4)if ( p2=p1+ 2 ){s1=s- (p1+p2)p3=sl/2-1; p4=p3+2:for ( j=0; j＜=prime_index-1; ++j )if ((5)){printf ( "%d* % d*36= (%d+ %d) + (%d+%d) \n", \ n,n, p1, p2, p3, p4 ) ;++t;}}}if ( t! = 0 ) printf ("%d\n", t )elseprintf ( "%d* %d*36=no so lution\n ", n, n ) ; }}}
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The other ______ is a special online edition that includes additional sports, lifestyle, and real estate pages.A. version
B. section
C. article
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4、4. interview __________
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