“编程与思维大赛”入门试题(C语言)

“编程与思维大赛”入门试题（C语言）程序设计题1：#include "stdio.h"#include "stdlib.h"#include <math.h>int main(){int N,**a,i,temp,j,k;printf("输入一个正整数：\n");scanf("%d",&N);a = (int **)malloc(N * sizeof(int *));for (i = 0; i < N; i++){scanf("%d",&a[i]);}for(i=0; i<N; i++){for(j=0; j<N-i-1; j++){if(a[j] < a[j+1]){temp = a[j];a[j] = a[j+1];a[j+1] = temp;}}}for (i = 0; i < N; i++){printf("%2d|%d\n",a[i],i);}printf("\n");for (i = 0; i < N; i++){if(a[i]<=i){printf("%d ",i);break;}}if(i==N){printf("%d",N);}return 0;}程序设计题2：#include "stdio.h"#include "stdlib.h"int main(){int m,n,i,j,k;float x[400],mse,**a,**b,p[20],q[20],temp,sum=0;printf("请输入行数m(<=20),列数n(<=20)\n");scanf("%d",&m);scanf("%d",&n);if(m>20||n>20){printf("error!");}a = (float **)malloc(m * sizeof(float *));for (i = 0; i < m; i++){a[i] = (float *)malloc(n * sizeof(float));for (j = 0; j < n; j++){scanf("%f", &a[i][j]);}}b = (float **)malloc(m * sizeof(float *)); for (i = 0; i < m; i++){b[i] = (float *)malloc(n * sizeof(float));for (j = 0; j < n; j++){scanf("%f", &b[i][j]);}}printf("数组A:\n");for (i = 0; i < m; i++){for (j = 0; j < n; j++){printf("%f ", a[i][j]);}printf("\n");}printf("数组B:\n");for (i = 0; i < m; i++){for (j = 0; j < n; j++){printf("%f ", b[i][j]);}printf("\n");k=1;for (i = 0; i <m; i++){for (j = 0; j < n; j++){p[k]=a[i][j];q[k]=b[i][j];k++;}}for(i=k;i<=20;i++){p[i]=0;q[i]=0;}for(i=0;i<m*n;i++){x[i]=(p[i+1]-q[i+1])*(p[i+1]-q[i+1]); }for(i=0; i<m*n; i++){for(j=0; j<k-i; j++){if(x[j] < x[j+1]){temp = x[j];x[j] = x[j+1];x[j+1] = temp;}}}for(i=0;i<m*n;i++){printf("%f\n",x[i]);}k=(int)(m*n*0.1);for(i=0;i<m*n-k;i++)x[i]=x[i+k];}for(i=0;i<m*n-k;i++){sum=sum+x[i];}printf("%0.5f\n",sum);mse=(float)(sum/(m*n-k));printf("%0.5f",mse);return 0;}程序设计题3：#include <stdio.h>int main() {int m, i, temp, num=1;int t[100];scanf("%d",&m);for(i=0;i<m;i++){scanf("%d",&t[i]);}if(t[0]>0){temp = 0;for(i=1;i<m;i++){if(t[i]<0 && -t[i]>t[0]) temp++;}num += temp;if(temp==0){printf("%d\n",num);return 0;}else{temp = 0;for(i=1;i<m;i++){if(t[i]>0 && t[i]<t[0]) temp++;}num += temp;}}else{temp = 0;for(i=1;i<m;i++) if(t[i]>0 && t[i]<-t[0]) temp++;num += temp;if(temp==0){printf("%d\n",num);return 0;}else{temp = 0;for(i=1;i<m;i++) if(t[i]<0 && t[i]<t[0]) temp++;num += temp;}}printf("%d\n",num);return 0;}思维拓展题1：思路：1.考虑分的第1份：分的数i不能大于n/k，因为大于n/k，就重复了。

2.从剩下的数n-i中分第2份（递归），剩下要分的份数为k-1，同样分的数i不能大于（n-i)/(k-1)……3.一直到k==1结束；4.考虑如何存储分法：使用数组，每分完第k份，就存到数组中对应的位置#include <iostream>using namespace std;#define Max 198// 得到分法int departNumber(int n, int k, int now) {// now表示当前至少从多少开始分，例如第一个应该是从1开始分if(k == 1) // 1份不用分了return 1;else {int count = 0;// F(n, k, now) = F(n - 1, k - 1, 1) + F(n - 2, k - 1, 2) + ... + F(n - n / k, k - 1, n / k)for(int i = now; i <= (n / k); i++)count = count + departNumber(n - i, k - 1, i); // 递归：把n-i分为k-1份，必须从i开始分return count;}}int printAllConditions(int n,int k,int start,int condition[Max],int index) {//n:整数，k:分成k份start:从start开始分类数组condition[Max]:存储分类的情况index:数组下标,从0开始if(k == 1) {condition[index] = n;for(int j = 0; j <= index; j++)cout << condition[j] << " ";cout << endl;return 1;} else {int sum = 0;for (int i = start; i <= (n / k); i++) {condition[index]=i;sum = sum + printAllConditions(n-i, k-1, i, condition, index+1);}return sum;}}int main(int argc, const char * argv[]) {// insert code here...int n, k;int conditions[Max];cout << "请输入整数n和份数k:" << endl;cin >> n >> k;cout << departNumber(n , k, 1) << endl;cout << printAllConditions(n, k, 1, conditions, 0) << "种" <<endl;return 0;}思维拓展题2：、思路：运用蒙特卡罗算法，用python语言，利用一到九个数，随机匹配计算，可以得到结果共有6种（Python）import randoma=[1,2,3,4,5,6,7,8,9]k=0n1s=[" "]n2s=[" "]def number(a,b,c,d):return 1000*eval(a)+100*eval(b)+10*eval(c)+eval(d)for i in range(5000000):a=["1","2","3","4","5","6","7","8","9"] #列表s1=random.sample(a,4) #从a里面随机抽取4个数n1=number(s1[0],s1[1],s1[2],s1[3])a.remove(s1[0])a.remove(s1[1])a.remove(s1[2])a.remove(s1[3])s2=random.sample(a,4)n2=number(s2[0],s2[1],s2[2],s2[3])if(n1-n2==2016):if(n1 not in n1s):k=k+1n1s.append(n1)n2s.append(n2) #排除相同元素if(i%1000000==0):print("一次")print(n1,n2)print(k)print(n1s)print(n2s)。