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1、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;i1000);num=fun(a,n);printf(“nnThe result:n”);for(i=0; iunsigned long fun(unsigned long n) u
2、nsigned long x=0,s,i; int t;s=n;/*found*i=0;/*found*while(s99999999|nunsigned 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|n#include #define M 5int fun(int
3、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;i1)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
4、现#include int Fib(int n);main() int i;for (i=0;ivoid main() fib(30); fib(int n) long int f100;int i;f0=0;f1=1;for(i=2;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 1.02
5、35Quotient 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*/,cu
6、rrent_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, while(money=1000, and multiple):); scanf(%d, printf(please input everyday net value in this week:n); for(i=0;i#include genlib
7、.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#define REDEMPTIONRATE 0.005 /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,
8、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
9、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 Increase Rate Current Value Current Payoff);for(i=0;i=4;i+) printf(%8d,(int)datei0);printf(%11.4f,datei1);printf(%14.2f%,datei2);printf(%15d,(int)(datei3+0.5); /为了四舍五入到整数所以+0.5printf(%16dn,(int)datei4);printf(Average Net Value in this week = %.5fn,sum/DAY);
