计算机二级填空题
全国计算机等级考试C 语言――填空题
1. s+= p->___1___; data p=p->___2___; next printf("\nsum=%d\n", fun(___3___)); head 2. n=len; k=___1___; i for(j=___2___; j>=0; j--) len ss[i][j]=____3____; '* 3.___1___fun(STU *std, char *num) STU if( strcmp(____2_____,num)==0 ) std[i].num return (___3____); std[i] 4. free(___1___); q q=p->___2___; next q=q->___3___; next
5.___1___fun(char ch) char if (ch>='0' &&___2____) ch
6. for(j=___1___; jn[j]) k=___2___; j
strcpy(ss[k],___3___); t 7. t = (a>b) ? (b>c? b :(a>c?c:___1___)) : ((a>c)?___2___ : ((b>c)?c:___3___)); 1. a 2. a 3.b 8. *n=____1_____; 0 p=____2____; p->next fun(head,____3____); &num 9. if((fs=fopen(source,___1___))==NULL) "r" while(!feof(___2___)) fs fputc(ch,___3___); ft 10. if(s[i]>='0' ___1___ s[i]data==ch) return ___2___; n k=fun(___3___); head,ch 13.void WriteText(FILE ___1___ ) *fw fputs(___2___,fw); fputs("\n",fw); str printf("%s",___3___); str 14. for(i=0; i
15.void fun(int ___1___ , int n) *a for (i=0; i='0'&&s[i]___3___) 27.#define OK(i, t, n) ((___1___%t==0) && (i/t
29. p = (n%2==0)?n/2:n/___1____; 2+1 a[___3___]= t; p+I t
0 i 1 j++ j "r" fs ft x t s[i] '9' n
a[i] 0 0 t[N][N] i=0;i
a[i] a[j] a[j] i t++ count STYPE FILE fp a[i] = a[p+___2___]; i p+i
30.void fun(int (*t)___1___ ) [N] ___2___ =t[i][j]+t[j][i]; t[i][j] ___3___ =0; t[j][i] 31. t=____1_____; t->next printf(" %d ",___2___); s->data free(___3___); t 32. f =___1___; 1.0 t *= (___2___)*x/n; -1 f += ___3___; t 33. ___1___; sum+=___2___ ; sum+= t[i][n-i-___3___] ; 34. t2[j]=s[i];___1___; for(i=0; idata=___1___; q=___2___; q->next=___3___; 41. if ((ch>='a') ___1___(ch___2___ *=a; modify(___3___,a);
sum=0 t[i][i] 1 j++ s[i]=t1[i] i
tt tt.score[i] std M
ss[i] n+j 1
STU score[i] &std
44.void fun(PERSON ___1___) std[ ] ___2___ temp; PERSON fun(___3___); std 45. fprintf(___1___,"%d %f\n",i,sqrt((double)i)); fp ___2___; fclose(fp) if((fp=fopen(___3___,"r"))==NULL) fname 46. for(i=0; i=0;___2___) t[j][N-1]=r[___3___]; 50. t = ___1___; t *= (-1.0)*x/___2___; while (____3____>= 1e-6); 51.___1___ fun(struct student a) for (i=0; i 0)
t=i i+2 '\0' *fw str str 0 j-- j x n fabs(t) struct student a.score[i] i tp FILE* fp ch '0' s++ sum [M] N '\0' *av i x[j] 1 2*i -1 struct student n-1 a[i].name,a[j].name
59. { p[__1__]=s[i]; j++;} for(i=0; idata __3__ q->data) 63. n=__1__; while(t=len) strcpy(__1__); for(i=len-n; i
j k p p->next q!=NULL p->next double f1 f2 h p->next > 0 x t++ NODE* q!=NULL r double f1 f2 feof(fp) ==sno sizeof(STU) a b.name score[i] N k-1 ss[i] *s s++ n 10 0 x t,s s[i] '\0' k N temp k
N
a[0][i] = __3__ ; 74. av=__1__; y[__2__]=x[i]; x[i]=-1;} if( x[i]!= __3__) y[j++]=x[i]; 75. av=__1__; for(i=0; iav) y[__3__]= x[i]; 76. t[j]=__1__; j++;} for(i=0; i
fscanf(__3__,"%s%s%s", str, str1, str2); 79. fp = fopen(__1__, "rb+"); fseek(fp, -1L*sizeof(STU), __2__); fwrite(__3__, sizeof(STU), 1, fp); 80. x=__1__/4; while(x __2__ e) x=__3__*(2*i+1)/(t*t); 81. for(i=0; i
fp = fopen(filename, __1__); if (s[i].sno __2__) __3__(s, sizeof(STU), N, fp);
a[k][i]
s/N j++ -1 0 x[i]/N j++ s[i] k '\0' 999 t/10 x FILE* floes(fp) fp filename SEEK_END &n 3.0 > (2*i-1) M i -1 ->sno ->name &t 0 || 1 struct student* a->score[i] a 0 n (t*t) n++ 0 s++ “rb” >s[j].sno fwrite
88. for(i=0; inext if (__2__) return; p->next==NULL p = q; q = __3__; r 91. for(i=0; i='A')&&(tt[i]
i++