【正文】 yn“)...、printf(”nIt is the...th “...。程序的運(yùn)行效果應(yīng)類似地如圖1所示，圖1中的200031是從鍵盤輸入的內(nèi)容。include int main(void){switch(month)printf(”Please input: yearmonthdayn“)。scanf(”%d%d%d“, amp。year, amp。month, amp。day)。int day, month, year, sum。return 0。if((iNum%5 == 0)amp。amp。(iNum%7 == 0)){ } else { } printf(”nNo.“)。printf(”nYes.“)。printf(”Please input an integer:“)。scanf(”%d“, amp。iNum)。{case 1:sum = 0。break。sum = 31。break。sum = 59。break。sum = 90。break。sum = 120。break。sum = 151。break。sum = 181。break。sum = 212。break。sum = 243。break。sum = 273。break。sum = 304。break。case 2: case 3: case 4: case 5: case 6: case 7: case 8: case 9: case 10: case 11:}} case 12: sum = 334。break。default: printf(”data error“)。break。sum += day。if((year%400==0 || year%4==0 amp。amp。 year%100!=0)amp。amp。 month2){ } printf(”nIt is the %dth “, sum)。return 0。sum++。輸入實(shí)型數(shù)據(jù)a,b，然后輸出a、b的值。程序的運(yùn)行效果應(yīng)類似地如圖1所示，,。從鍵盤輸入x,y,z的值，編寫程序輸出以下表達(dá)式的值： x+z%3*(int)(x+y)%2/4程序的運(yùn)行效果應(yīng)類似地如圖1所示，,7是從鍵盤輸入的內(nèi)容。從鍵盤輸入一日期，年月日之間以“”分隔，并以同樣的形式但以“/”作分隔符輸出。程序的運(yùn)行效果應(yīng)類似地如圖1所示，圖1中的2009129是從鍵盤輸入的內(nèi)容。include int main(void){} 輸入三角形的三邊長a、b、c(邊長可以是小數(shù))，求三角形面積area，并輸出。如果輸入的三邊構(gòu)不成三角形，應(yīng)給出“data error”的信息提示。注：根據(jù)“海倫－秦九韶”公式，area＝√p(pa)(pb)(pc)，其中p＝(a+b+c)/2。編程可用素材：printf(”nplease input triange sides:“)...、printf(”ndata errorn“)...、printf(”narea=...n“...。程序的運(yùn)行效果應(yīng)類似地如圖1和圖2所示，圖1中的3,4,5和圖2中的3,4,8是從鍵盤輸入的內(nèi)容。include include int main(void){if(bianA} { } ||(bianA+bianB } return 0。pABC = / 2 *(bianA + bianB + bianC)。area = sqrt(pABC *(pABCbianB)*(pABC500P115 25 * P115 25 / P115 25 %P115 25 x2P115 100 +和P115 100 + 330 =是從命令行輸入的內(nèi)容(注：圖中的Pxxxxx表示隨考生題號變換的內(nèi)容，在本套試卷中，請考生在閱讀時將圖中的Pxxxxx視作P115)。include include int main(int argc, char *argv[]){switch(argv[2][0]){case 39。+39。 :result = num1 + num2。break。result = num1hn。sn = sn + hn。hn = hn / 2。sn = sn + hn。printf(”Please input n:“)。scanf(”%d“, amp。n)。int i, n。double sn, hn。} return 0。求s=a+aa+aaa+aaaa+aa...a的值，其中a是一個數(shù)字（可取1～9之間的一個值）。例如2+22+222+2222+22222（此時共有5個數(shù)相加），其中a值和有幾個數(shù)相加由鍵盤輸入控制。注意s的值有可能超出int的范圍，編程可用素材：printf(”Please input a,n: “)...、printf(”a+aa+...=...n“...。程序的運(yùn)行效果應(yīng)類似地如圖1所示，圖1中的2,3是從鍵盤輸入的內(nèi)容。includeint main(void){} 輸入兩個正整數(shù)m和n，求其最大公約數(shù)和最小公倍數(shù)。注：最大公約數(shù)也稱最大公因子，指某幾個整數(shù)共有因子中最大的一個；兩個整數(shù)公有的倍數(shù)稱為它們的公倍數(shù)，其中最小的一個正整數(shù)稱為它們兩個的最小公倍數(shù)。編程可用素材：printf(”please input two integer numbers: “)...、printf(”nthe greatest mon diprintf(“a+aa+...=%.0fn”, sn)。return 0。while(count } tn = tn + valA。sn = sn + tn。valA = valA * 10。count++。printf(“Please input a,n: ”)。scanf(“%lf,%d”, amp。valA, amp。n)。int n, count=1。double valA, sn=0, tn=0。visor is...n“...、printf(”the least mon multiple is...n“...。程序的運(yùn)行效果應(yīng)類似地如圖1所示，圖1中的35 15是從鍵盤輸入的內(nèi)容。include includeint main(void){} 某班有40位同學(xué)參加考試，成績（整數(shù)）從鍵盤輸入，求全班最高分、最低分以及平均分，并統(tǒng)計該班同學(xué)的考試及格率。編程可用素材：printf(”n請輸入40位同學(xué)的成績：“)...、printf(”n最高分：...最低分：...平均分：...及格率：...。return 0。printf(“nthe greatest mon divisor is %dn”, zdGys)。printf(“the least mon multiple is %dn”, zxGbs)。zdGys = min(m, n)。while(m % zdGys!= 0||n%zdGys!= 0){ }zxGbs = max(m, n)。while(zxGbs % m!=0||zxGbs % n!= 0){ } zxGbs++。zdGys。printf(“please input two integer numbers: ”)。scanf(“%d%d”, amp。m, amp。n)。int m, n, zdGys, zxGbs。程序的運(yùn)行效果應(yīng)類似地如圖1所示，圖1中的99 81 71 81 77 94 100 67 66 44 75 49 47 45 65 74 73 74 63 69 72 77 65 79 84 73 46 62 68 42 75 62 65 66 62 69 44 62 84 77是從鍵盤輸入的內(nèi)容。include define SIZE 40int main(void){/*求最大、最小、分?jǐn)?shù)總和，統(tǒng)計及格人數(shù)*/ max = scores[0]。min = scores[0]。aver = 0。count = 0。for(i = 0。i SIZE。i++){if(scores[i] max){ } else if(scores[i] min){ min = scores[i]。max = scores[i]。/*從鍵盤接收數(shù)據(jù)*/ printf(“n請輸入%d位同學(xué)的成績：”, SIZE)。for(i = 0。i SIZE。i++){ } scanf(“%d”, amp。scores[i])。int scores[SIZE], i。int max, min, count。double aver, rate。}aver += scores[i]。if(scores[i] = 60){count++。} }/*求平均分和及格率*/ aver = aver / 40。rate =(double)count / 40。/*輸出結(jié)果*/ printf(“n最高分:%dn最低分: n”, max, min, aver, rate*100)。return 0。}%dn平均分: %.1fn及格率: %.0f%%第三篇：成都信息工程大學(xué) 專利申請承諾書專利申請承諾書編號：擬申請的專利名稱：申請專利類型：我校著名為第名職務(wù)發(fā)明人：所在部門：合作單位：我保證提供的任何有關(guān)知識和技術(shù)均不侵犯任何人的知識產(chǎn)權(quán)，不得將屬于校方的知識產(chǎn)權(quán)私自交付他人。若因此產(chǎn)生的法律糾紛，本人愿承擔(dān)全部責(zé)任。承諾人:年 月 日第四篇：成都信息工程大學(xué) 學(xué)位證英文版3rd, July, 2009Degree CertificateThis is to certify that xxx, male, born on 3rd, February, 1986 , has been studied in Chengdu University of Information Technology from September, 20xx to July, 20xx majoring in Tourism has pleted all the courses prescribed in the Bachelor Program, and has passed all the exams necessary for fulfilled the requirements stipulated by the Academic Degree Regulation of the People’s Republic of China, he is awarded the Bachelor Degree of Management(Science, Engineering, Arts).Lai TingQianChairman of the Degree AppraisalCommittee of Chengdu University of Information Technology: 1ST, July, 2009Certificate No.: 10621xxxxxxxxxxx第五篇：成都信息工程大學(xué) 在讀證明英文版XXth, January, 2010Studying CertificateThis is to certify XXX, female, born on xxth, March, 19xx, the Student , has been majored in Tourism Management(fouryear Bachelor Degree)at Chengdu University of Information Technology since September will get the Bachelor in July 1st, 20xx if she passes the examinations in the required hereby version has been confirmed by Teaching Affairs Office Chengdu University of Information TechnologyFax: 02885966503