近几年三级网络技术上机高频题(必考).doc

int i, j, half, hun, ten, data;for (i=101; i1000; i+)hun = i/100;ten = i%100/10;data = i%10;if (hun = (ten+data)%10)half = i/2;for (j=2; j= half)cnt+; sum += i;数的按位分离及合并int i, j;int a1, a2, a3, a4, k;for (i=0; iMAX; i+)a1 = ai/1000;a2 = ai%1000/100;a3 = ai%100/10;a4 = ai%10;if (a1 = a2) & (a2 = a3) & (a3 = a4) & (ai%2 = 0)bcnt = ai;cnt+;for (i=0; icnt-1; i+)for (j=i+1; j bj)k = bi;bi = bj;bj = k;数位分解后按特定关键字进行排序int i, j, data;for (i=0; i199; i+)for (j=i+1; j aaj%1000)data = aai;aai = aaj;aaj = data;else if (aai%1000 = aaj%1000)if (aai aaj)data = aai;aai = aaj;aaj = data;for (i=0; i10; i+)bbi = aai;相邻数的筛选统计int i, j, flag = 0;int k;for (i=0; iMAX-5; i+)for (j=i+1; j aj)flag = 1;elseflag = 0;break;if (flag=1 & ai%2)bcnt = ai;cnt+;for (i=0; icnt-1; i+)for (j=i+1; j bj)k = bi;bi = bj;bj = k;数值筛选并统计int i, thou, hun, ten, data;int ab;long sum = 0;for (i=0; i 0)totNum+;thou = xxi/1000;hun = xxi%1000/100;ten = xxi%100/10;data = xxi%10;ab = thou+hun+ten+data;if (ab%2 = 0)totCnt+;sum = sum+xxi;totPjz = (double)sum/totCnt;方差计算for (i=0; i100; i+)for (j=0; j10; j+)fscanf(fp, %d, &xxi*10+j);fscanf(fp, n);if (feof(fp)break;按条件筛选并替换字符int i;char *pf;for (i=0; imaxline; i+)pf = xxi;while (*pf != 0)if (*pf*11%25632)*pf = *pf*11%256;pf+;字符串特殊排序int i, j, k, strl;char ch;for (i=0; i20; i+)strl = strlen(xxi);for (j=1; jstrl-2; j=j+2)for (k=j+2; k xxik)ch = xxij;xxij = xxik;xxik = ch;按结构体成员进行多关键字排序int i, j;PRO xy;for (i=0; i99; i+)for (j=i+1; j 0)xy = selli;sell i = sellj;sellj = xy;else if (strcmp(selli.mc, sellj.mc) = 0)if (selli.je sellj.je)xy = selli;selli = sellj;sellj = xy;结构体筛选并排序int i, cnt = 0, j;data ch;for (i=0; i aai.x1+aai.x3)bbcnt = aai;cnt+;for (i=0; icnt-1; i+)for (j=i+1; jcnt; j+)if (bbi.x2+bbi.x3 bbj.x2+bbj.x3)ch = bbi;bbi = bbj;bbj = ch;return cnt;选票问题int i, count, j;char *pf;for (i=0; i10; i+)yyi = 0;for (i=0; i 5)for (j=0; j10; j+)yyj += xxij-0;出圈问题int i, j, s1, w;s1 = s;for (i=1; i=2; i-)s1 = (s1+m-1)%i;if (s1 = 0)s1 = i;w = ps1-1;for (j=s1; j=i-1; j+)pj-1 = pj;pi-1 = w;判断回文数int i, strl, half;char xy20;ltoa(n, xy, 10);strl = strlen(xy);half = strl/2;for (i=0; i= half)return 1;elsereturn 0;以地推关系求数列值int f1 = 0, f2 = 1, fn;fn = f1+f2;while (fn = t)f1 = f2;f2 = fn;fn = f1+f2;return fn;用迭代法求方程的根float X0, X1 = 0.0;while (1)X0 = X1;X1 = (float)cos(X0);if (fabs(X0-X1) 1e-6)break; return X1;