2020浙江工商计算机机试1

1grade band count

#include <stdio.h>
int main(void) {
    int a[20];
    int i,j,n,grade,excellent=0,pass=0,fail=0;
    printf("please input n equal to:");
    scanf("%d",&n);
    printf("a[n] equal to:");
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
        if(a[i]<=0){
            printf("exit");
            break;
        }
    }
    for(i=0;i<n;i++){
        printf("%d\t",a[i]);
    }
    for(j=0;j<n;j++){
        if(a[j]>=85){
            excellent++;
        }
        else if(a[j]>=60){
            pass++;
        }
        else{
            fail++;
        }
    }
    printf("excellent=%d,pass=%d,fail=%d",excellent,pass,fail);
    return 0;
}

2chicken and rabbit cage

#include <stdio.h>

int main(void) {
    int n;
    printf("there are n feet in the cage\n");
    scanf("%d",&n);
    printf("there are at most %d animals in the cage\n",n/2);
    if(n%4==0){
        printf("there are at least %d animals in the cage\n",n/4);
    }
    else{
        printf("there are at least %d animals in the cage\n",(n-2)/4+1);
    }
    return 0;
}

3compound interest method

这种利率的问题居然能利用二分法来解决，这也太离谱了。
就是利用这种无穷的思想，硬生生找一个利率出来，找不到就换一个利率，反正利率肯定大于0小于1。
不断逼近真正的利率，这种思想才是最可贵的。如果sum大于0，就表明还在欠钱，难难难，这种思维就不是人的思维。

#include <stdio.h>
#include <math.h>
int main(){
	double a;
	int month;
	double c;
	scanf("%lf %d %lf",&a,&month,&c);
	double sum = a,low = 0,high = 1,mid;//月利率介于0%和100%之间，sum为欠的钱
	while(fabs(sum) >= 0.00001){
		sum = a;
		mid = (low+high)/2;
		for(int i = 0;i < month;i++){
			sum = sum*(1+mid)-c;
		}
		if(sum > 0)//sum大于0说明利率太高，按每月还的钱来算还会欠钱
			high = mid;
		else 
			low = mid;
	}
	printf("%.3lf%%\n",mid*100);
	return 0;
}

4考场如战场，我不但要写得好，还要写的快，这样才能拿分；时常练习很重要

#include <stdio.h>
//判断一个数是否是守型数
int  main(void) {
    int m,k,n;
    int i=1,j=1;
    printf("要判断的正整数\n");
    scanf("%d",&n);
    m=n;
    while(m/10>=1){
        m=m/10;
        i++;
    }
    printf("n是%d位数",i);
    k = n*n;
    while(k/10>=1){
        k=k/10;
        j++;
    }
    printf("n*n是%d位数",j);
    if(n<=3){
        switch(n){
            case 1:
            printf("yes");
            break;
            case 2:
            printf("no");
            break;
            case 3:
            printf("no");
            break;    
        }
    }
    else{
        if(n*n%(j-i)*10==n){
            printf("yes");
        }
        else{
            printf("no");
        }
    }

    return 0;
}