C语言实验三参考答案-2012秋.doc

实验5、函数1 程序填空给定程序中函数fun的功能是：找出100n（不大于1000）之间的三位数字相等的所有整数，把这些整数放在s所指的数组中，整数的个数作为函数值返回。在程序的下划线处填入正确的内容并把下划线删除，使程序得出正确的结果即可。程序填空如下：#include #define N 100int fun(int *s,int n) int i,j,k,a,b,c;j=0;for(i=100;in;i+) /*found*/k= i ;a=k%10; k/=10;b=k%10; k/=10;/*found*/c= k ;if(a=b&a=c) sj+=i;return j;main() int aN,n,num=0,i;do printf(“nEnter n(1000);num=fun(a,n);printf(“nnThe result:n”);for(i=0; inum; i+) printf(“%5d”,ai);printf(“nn”);2 程序改错给定程序中函数fun的功能是：将形参n中各位上为偶数的数取出，并按原来从高位到低位的顺序组成一个新的数，并作为函数值返回。例如，从主函数输入一个整数27638496，函数返回值为26846。请改正函数fun中指定部位的错误，使它能够得出正确的结果。注意：不要改动main函数，不得增行或删行，也不得更改程序的结构。需改正程序如下：#include unsigned long fun(unsigned long n) unsigned long x=0,s,i; int t;s=n;/*found*i=0;/*found*while(s99999999|n0)printf(“Please input(0n100000000):“);scanf(“%ld”,&n); printf(“nThe result is:%ldn”,fun(n);正确的程序：#include unsigned long fun(unsigned long n) unsigned long x=0,s,i; int t;s=n;/*found*/i=1;/*found*/while(s0) t=s%10;if(t%2=0)/*found*/x=x+t*i; i=i*10;s=s/10;return x;main() unsigned long n=1;while(n99999999|n0)printf(“Please input(0n100000000):“);scanf(“%ld”,&n); printf(“nThe result is:%ldn”,fun(n);3、编写程序（1）给定程序中函数fun的功能是：输出M行M列的整数方阵，然后求两条对角线上的元素之和，要求返回此和数。同时，在函数中需完整输出整数方阵和对角线上元素之和。因此，编写程序时需注意方阵元素的排列顺序和求对角线元素之和时不要漏项。另外，两条对角线交叉点上的一个值允许重复计算。注意：切勿改动主函数main和其他函数中的任何内容，仅在函数fun的花括号中填入编写的语句。程序如下：#include #include #define M 5int fun(int n,int xxMM)main()int aaMM = 1,2,3,4,5,4,3,2,1,0,6,7,8,9,0,9,8,7,6,5,3,4,5,6,7;printf(“nThe sum of all elements on 2 diagnals is %d.“,fun(M,aa);参考程序如下：#include #include #define M 5int fun(int n,int xxMM) int i,j,sum=0;for(i=0;in;i+)for(j=0;jn;j+)printf(%d ,xxij);sum+=xxii+xxn-i-1i;printf(n);return sum;main()int aaMM = 1,2,3,4,5,4,3,2,1,0,6,7,8,9,0, 9,8,7,6,5,3,4,5,6,7;printf(“nThe sum of all elements on 2 diagnals is %d.”,fun(M,aa);也可以分成多个for循环语句来写：int fun(int n,int xxMM)int i,j;int num1 = 0, num2 = 0,allnum;for(i=0; in; i+)for(j=0;jn;j+)printf(%5d,xxij);printf(n);for(i = 0; i n; i+)num1 += xxii;printf(主对角线的和是: %dn,num1);for(i = 0; i 1)从上述公式可以看出：斐波那契数列的第1个数是0，第2个数是1，从第3个数开始，以后每个数都是前两个数之和。解：方法一：用递归法实现#include int Fib(int n);main()int i;for (i=0;i30;i+)printf(F(%d)=%dn,i,Fib(i);int Fib(int n)if (n=0)return 0;else if (n=1)return 1;elsereturn Fib(n-1)+Fib(n-2);方法二：用递推法实现#includevoid main() fib(30); fib(int n) long int f100;int i;f0=0;f1=1;for(i=2;in;i+) fi=fi-2+fi-1;for(i=0;i1000, and multiple): 15001500 is not multiple of 1000.Please input your money(1000, and multiple): 10000Please input everyday net value in this week:0.9872 0.9935 1.0102 0.9905 1.0235Quotient Net Value Increase Rate Current Value Current Payoff9850 0.9872 0.00% 9724 -3259850 0.9935 0.64% 9786 -2639850 1.0102 1.68% 9950 -999850 0.9905 -1.95% 9756 -2929850 1.0235 3.33% 10081 31Average Net Value in this week = 1.00098方法、用一维数组实现：#include void main() int money,i,quotient/*基金份额*/,current_value5/*用户当前净值*/,current_payoff5/*浮动盈亏*/; float net_value5/*当日每份基金净值*/,increase_rate5/*每日净值增长率*/,sum=0; printf(Please input your money(=1000, and multiple):); scanf(%d,&money); while(money1000|(money%1000)!=0) printf(%d is not multiple of 1000 or =1000, and multiple):); scanf(%d,&money); printf(please input everyday net value in this week:n); for(i=0;i5;i+) scanf(%f, &net_valuei); quotient=(money*0.985)/1; for(i=0;i5;i+) if(i=0) current_valuei=quotient*net_valuei; increase_ratei=(net_valuei-1); current_payoffi=(float)(current_valuei*(1-0.005)-money); else current_valuei=quotient*net_valuei; increase_ratei=(net_valuei-net_valuei-1)/net_valuei-1; current_payoffi=(float)(current_valuei*(1-0.005)-money); printf(Quotient Net Value Increase Rate Current Value Current Payoffn);for(i=0;i5;i+) sum=sum+net_valuei; printf( %-8d %-3.4f %10.2f% %12d %15dn,quotient,net_valuei,increase_ratei*100,current_valuei,current_payoffi);printf(Average Net Value in this week = %fn,sum/5);方法、用二维数组实现：#include #include genlib.h#include simpio.h#include strlib.h#define PREVALUE 1/成立时每份面值#define PURCHASERATE 0.015/申购费率#defineREDEMPTIONRATE0.005/赎回费率#define DAY 5/工作日void main(void)int amount,i;double dateDAY5;double sum=0.0;printf(Please input your money(=1000, and multiple): );amount=GetInteger();while(amount=1000, and multiple): );amount=GetInteger();printf(Please input everyday net value in this week: );for(i=0;i5;i+)datei0=amount*(1-PURCHASERATE)/PREVALUE;/Quotient-基金份额datei1=GetReal();/Net value-当日每份基金净值datei3=datei0*datei1;/Current Value-用户当前净值datei4=datei3*(1-REDEMPTIONRATE)-amount;/Current Payoff-浮动盈亏if(datei40)datei4=datei4-0.5;/当Current Payoff为负时-0.5以便四舍五入elsedatei4=datei4+0.5;/当Current Payoff为正时+0.5以便四舍五入sum=sum+datei1;date02=-1.28;/由于上周最后一个交易的当日每份基金净值未知，所以第一日的每日净值增长率由程序定义for(i=1;i=4;i+)datei2=(datei1-datei-11)/datei-11*100; /Increase Rate-每日净值增长率printf(Quotient Net Value Increas