二级C上机编程题课堂讲解.docx

15void fun ( int ttMN, int ppN )int min, i, j, k = 0;for(j = 0; j N; j+)min = tt0j;for(i = 1; i M; i+)if(ttij min) min = ttij;ppk+ = min;30double fun(double x9)double sum = 0;int i;for(i = 0; i 8; i+)sum += sqrt(xi + xi + 1) / 2); return sum;42void fun(int a, int n , int *max, int *d )int m, dd, i;m = a0;dd = 0;for(i = 1; i m)m = ai;dd = i;*max = m;*d = dd;43int fun (int aM)int max, i, j;max = a00;for(i = 0; i 2; i +)for(j = 0; j max) max = aij;return max;48void fun ( int aN, int m )int i, j;for(i = 0; i N; i+)for(j = i; j N; j+)aij *= m;49double fun ( int wN )double ave = 0;int i;for(i = 0; i N; i+)ave += w0i + wN - 1i + wi0 + wiN - 1;ave = ave - w00 - wN - 10 - w0N - 1 - wN - 1N - 1;ave /= 4 * N - 4;return ave;52void fun(int array33)int i, j;for(i = 0; i 3; i+)for(j = i; j 3; j+) /特别注意j的初值int t = arrayij;arrayij = arrayji;arrayji = t;77void fun(int (*s)10, int *b, int *n, int mm, int nn)int i, j;for(i = 0; i mm; i+)for(j = 0; j nn; j+)bi * nn + j = sij;*n = mm * nn;80void fun(int *w, int p, int n)int aN, i, k = 0;for(i = p + 1; i n; i+)ak+ = wi;for(i = 0; i = p; i+)ak+ = wi;for(i = 0; i n; i+)wi = ai;83int fun(int a, int n)int cnt = 0, bN, i, j;bcnt+ = a0;for(i = 1; i n; i+)if(ai != bcnt - 1) bnct+ = ai; for(i = 0; i cnt; i+)ai = bi;return cnt;84void fun( int *a, int *b)int i;for(i = 0; i M; i+)bi = 0;for(i = 0; i = 100)b10+;elsebai / 10+;87int fun( int *s, int x)int i;for(i = 0; i N; i+)if(x = si)return i;return -1;7void fun ( int m, int *a , int *n )int x, k = 0;for(x = 1; x = m; x+)if(x % 7 = 0 | x % 11 = 0)ak+ = x;*n = k;26int fun(int n)int sum = 0, yz;for(yz = 2; yz n; yz+)if(n % yz = 0) sum += yz;return sum;34double fun(int n)double sum = 0;int ss, i;for(ss = 3; ss = n; ss+)for(i = 2; i ss; i+)if(ss % i = 0) break;if(i = ss) sum += sqrt(ss);return sum;51return (int)(h * 100 + 0.5) / 100.0;53int fun(int lim, int aaMAX) /* 以下代码仅供参考 */ int i,j,k=0; /* 其中变量k用于统计素数个数 */ for(i=2;i=lim;i+) /* 以下找出小于或等于lim的素数存入aa数组中并统计素数个数 */ for(j = 2; j i; j+) if(i % j = 0) break; if(j = i) aak+ = i; return k;6#include char *fun ( char *s, char *t)if(strlen(s) = strlen(t)return s;elsereturn t;39#include void fun(char p1, char p2)strcat(p1, p2);40#include long fun ( char *p) long x=0; x = atol(p); return x;54void fun(char(*a)81,int num,char *max) /* 以下代码仅供参考 */ int i,k=0,len, maxlen; /* k为a数组中最长串所在元素的下标，初始为0，maxlen为其串长 */ maxlen=strlen(ak); for(i=1;i strlen(ak) k = i; /此句为答案 *max=ak;56void fun( char *ss )int i;for(i = 0; ssi != 0; i+) /关键语句：ssi != 0if(i % 2 = 1 & ssi = a & ssi = z)ssi -= 32;76int fun(char *str)int n, i, res = 1;n = strlen(str);for(i = 0; i = 0 & *s = 9) cnt+;s+;return cnt;88#include void fun( char *a , char *s)strcpy(a, s);89#include long fun( char *s )return atol(s);4void fun(char *s, char t)int i, k = 0;for(i = 0; si; i+)if(i % 2 = 0 & si % 2 = 0)tk+ = si;tk = 0; /very important!8void fun ( char *ss )int i;for(i = 0; ssi;i+)if(i % 2 = 1 & ssi = a & ssi = z)ssi -= 32;44void fun(char *s, char t)int i, k = 0;for(i = 0; si; i+)if(i % 2 = 1 & si % 2 = 1)tk+ = si;tk = 0;58int fun(int score, int m, int below)double ave = 0;int k = 0;for(i = 0; i m; i+)ave += scorei;ave /= m;for(i = 0; i m; i+)if(scorei ave)belowk+ = scorei;return k;27double fun(int n)double s = 1, t = 1;int i;for(i = 1; i = 0.000001);return s;29double fun(double x10)double s = 0, x1 = 0;int i;for(i = 0; i 10; i+)x1 += xi;x1 /= 10; for(i = 0; i 10; i+)s += (xi - x1) * (xi - x1);s /= 10;s = sqrt(s);return s;35double fun(int n)double s = 0, t = 0;int i;for(i = 1; i = eps); s *= 2;return s;62float fun(int m, int n)float fz = 1, fm = 1, p;int i;for(i = n + 1; i = m; i+)fz *= i;for(i = 1; i = m - n; i+)fm *= i;p = fz / fm;return p;2int fun( STREC *a, STREC *b )STREC min;int i, k = 0;min = a0;for(i = 1; i N; i+)if(ai.s min.s)min = ai;for(i = 0; i N; i+)if(ai.s = min.s)bk+ = ai;return k;12void fun( STU a, STU *s )STU max;int i;max = a0;for(i = 1; i max.s) max = ai;*s = max;21void fun(STREC *a)int j;a-ave = 0;for(j = 0; j ave += a-sj; a-ave /= N;22double fun( STREC *a, STREC *b, int *n)double ave = 0;int i, k = 0;for(i = 0; i N; i+)ave += ai.s;ave /= N;for(i = 0; i N; i+)if(ai.s ave)bk+ = ai;*n = k;return ave;24STREC fun( STREC *a, char *b ) int i; STREC t = 0, -1; for(i = 0; i next;while(p != NULL)ave += p-s;p = p-next;ave /= N;return ave;37void fun( STREC a ) STREC tmp; int i,j; for(i = 0; i N; i+) for(j