信息学竞赛典型例题程序汇集

信息学竞赛复习1.斐波拉契数列（1000000以内的数字）#include “stdio.h”#include “stdlib.h”void main(){long fib1=1,fib2=1,fib=0;printf("1000000以内的数字:\n\n");printf("%d\n%d\n",fib1,fib2);while (fib<1000000){fib=fib1+fib2;fib1=fib2;fib2=fib;printf( "%d \n", fib);}system("pause");}2.约瑟夫问题（1）约瑟夫问题_数组#include "stdio.h"#include "stdlib.h"#define Length 100void main(){int i,j;int len; //输入的总节点数int n,m; //输入的每次数几个int remain; //剩下的节点数int current; //当前数到那个数字int a[Length];for(i=0;i<Length;i++){a[i]=1;}printf("请输入约瑟夫问题节点总数：");len=41; //scanf("%d",&len);printf("请输入每次数的节点数字：");n=3; //scanf("%d",&n);remain=len; //剩下的节点数current=0; //当前数到那个数字m=0; //计数到n，m=n时清零//从1开始数//循环到剩下的节点数为0printf("出列的节点的编号依次为：");while(remain>0){current++;while(current>len){current-=len;}if(a[current]==1){m++;if(m==n){a[current]=0;remain--;printf("%d, ",current);m=0;}}}system("pause");}（2）约瑟夫问题_数组环#include "stdio.h"#include "stdlib.h"void main(){int a[100];int i,len,n; //i循环 len总数 n每次数几个int cur,m,remain; //cur当前是哪个 m 计数 remain 剩下几个//int t1,t2;for(i=0;i<100;i++){a[i]=i+1;}printf("请输入约瑟夫问题节点总数：");len=41; //scanf("%d",&len);printf("请输入每次数的节点数字：");n=3; //scanf("%d",&n);a[len]=1; //首尾相连remain=len; //开始前，剩余数为总数cur=1; //当前从第1个开始m=1; //计数从1开始while(remain>0){cur=a[cur];m++;if(m==n-1){printf("%d, ",a[cur]);//t1=a[cur];//t2=a[a[cur]];a[cur]=a[a[cur]];remain--;m=0;}}system("pause");}（3）约瑟夫问题_链表#include "stdio.h"#include "stdlib.h"struct people{int num;struct people *next;};typedef struct people node;node *create(int m){int i;node *h,*p1,*p2;h=p1=p2=(node *)malloc(sizeof(node));h->num=1;for(i=1;i<m;i++){p1=(node *)malloc(sizeof(node));p2->next=p1;p1->num=i+1;p2=p1;}p1->next=h;return h;}node *findout(node *tp,int n){int i;node *p;p=tp;for(i=1;i<n-1;i++){p=p->next;}return p;}node *moveaway(node *tp){node *c,*p1,*p2;c=tp;p1=c->next;p2=p1->next;c->next=p2;printf("%d, ",p1->num);free(p1);return p2;}void main(){node *p;int len,i,n;printf("请输入约瑟夫问题节点总数：");len=41; //scanf("%d",&len);printf("请输入每次数的节点数字：");n=3; //scanf("%d",&n);p=create(len);for(i=1;i<=len;i++){p=moveaway(findout(p,n));}system("pause");}（4）约瑟夫问题_双向链表#include "stdio.h"#include "stdlib.h"struct people{int num;struct people *next;struct people *prior;};typedef struct people node;node *create(int m){int i;node *h,*p1,*p2;h=p1=p2=(node *)malloc(sizeof(node));h->num=1;for(i=1;i<m;i++){p1=(node *)malloc(sizeof(node));p2->next=p1;p1->prior=p2;p1->num=i+1;p2=p1;}p1->next=h;h->prior=p1;return h;}node *findout(node *tp,int n){int i;node *p;p=tp;for(i=1;i<n;i++){p=p->next;}return p;}node *moveaway(node *tp){node *c,*p1,*p2;c=tp;p1=c->prior;p2=c->next;p1->next=p2;p2->prior=p1;printf("%d, ",c->num);free(c);return p2;}void main(){node *p;int len,i,n;printf("请输入约瑟夫问题节点总数：");len=41;//scanf("%d",&len);printf("请输入每次数的节点数字：");n=3;//scanf("%d",&n);p=create(len);for(i=1;i<=len;i++){p=moveaway(findout(p,n));}system("pause");}3.数字7 和5（1）统计含有数字7，但不能被7整除的5位整数的个数#include <stdio.h>#include <stdlib.h>void main(){int sum=0;//统计个数int b;//起数int a[6];int have7;int j;int tempb;for(b=10000;b<100000;b++){//分离数字到数组tempb=b;for(j=0;j<5;j++){a[5-j]=tempb-(tempb/10)*10;tempb=tempb/10;}//判断是否含7have7=0;for(j=1;j<=5;j++){if(a[j]==7){have7++;}}if(have7>0 && b%7!=0){//printf("%d\n",b);sum++;}}printf("sum=%d\n",sum);system("pause");}（2）统计含有2个数字7，但不能被7整除的5位整数的个数#include <stdio.h>#include <stdlib.h>void main(){int sum=0;//统计个数int b;//起数int a[6];int have7;int j;int tempb;for(b=10000;b<100000;b++){//分离数字到数组tempb=b;for(j=0;j<5;j++){a[5-j]=tempb-(tempb/10)*10;tempb=tempb/10;}//判断是否含7have7=0;for(j=1;j<=5;j++){if(a[j]==7){have7++;}}if(have7==2 && b%7!=0){printf("%d\n",b);sum++;}}printf("sum=%d\n",sum);system("pause");}4.百钱买百鸡：鸡翁一，值钱五；鸡母一，值钱三；鸡雏三，值钱一；百钱买百鸡，问鸡翁、鸡母、鸡雏各几何？（1）#include <stdio.h>#include <stdlib.h>void main(){int i,j,k;for(i=1;i<20;i++){for(j=1;j<33;j++){k=100-i-j;if(i*15+j*9+k==300){printf("鸡翁%d,鸡母%d,鸡雏%d\n",i,j,k);}}}system("pause");}（2）#include <stdio.h>#include <stdlib.h>void main(){int i,j,k;for(i=1;i<20;i++){for(j=1;j<33;j++){k=(100-i*5-j*3)*3;if(i+j+k==100){printf("鸡翁%d,鸡母%d,鸡雏%d\n",i,j,k);}}}system("pause");}5.逆序乘积式（1）两位数逆序乘积式#include <stdio.h>void main(){int sum=0;int a,b,c,d;int i,j,ir,jr,m,n;int arr[5];int flag;for(i=10;i<98;i++){arr[1]=a=i/10;arr[2]=b=i-a*10;for(j=i+1;j<98;j++){arr[3]=c=j/10;arr[4]=d=j-c*10;//检查有没有相同数字flag=0;for(m=1;m<4;m++){for(n=m+1;n<5;n++){if(arr[m]==arr[n]){flag=1;break;}}}//检查完//右侧的逆序ir=b*10+a;jr=d*10+c;if(flag==0&&i*j==ir*jr&&i<j&&i<ir&&i<jr){sum++;printf("%d:\ti=%d\t,j=%d\t,ir=%d\t,jr=%d\n",sum,i,j,ir,jr);}}}}（2）三位数逆序乘积式#include <stdio.h>void main(){int sum=0;int a,b,c,d,e,f;int i,j,ir,jr,m,n;int arr[7];int flag;for(i=100;i<987;i++){arr[1]=a=i/100;arr[2]=b=i/10-a*10;arr[3]=c=i-a*100-b*10;for(j=i+1;j<987;j++){arr[4]=d=j/100;arr[5]=e=j/10-d*10;arr[6]=f=j-d*100-e*10;//检查有没有相同数字flag=0;for(m=1;m<6;m++){for(n=m+1;n<7;n++){if(arr[m]==arr[n]){flag=1;break;}}}//检查完//右侧的逆序ir=c*100+b*10+a;jr=f*100+e*10+d;if(flag==0&&i*j==ir*jr&&i<j&&i<jr){sum++;printf("%d:\ti=%d\t,j=%d\t,ir=%d\t,jr=%d\n",sum,i,j,ir,jr);}}}}6.双和数组（1）普通解法#include <stdio.h>void main(){int s;int a,b,c,d,e,f;int arr[7];int flag,i,j;int sum=0;for (s=11;s<=100;s++){printf("s=%d:\n",s);for(a=1;a<=s-2;a++){for(b=a+1;b<=s-1;b++){for(d=a+1;d<=s-2;d++){for(e=d+1;e<=s-1;e++){c=s-a-b;f=s-d-e;//检查是否有重复数字arr[1]=a;arr[2]=b;arr[3]=c;arr[4]=d;arr[5]=e;arr[6]=f;if(a<b&&b<c&&d<e&&e<f){flag=0;for(i=1;i<7;i++){for(j=i+1;j<7;j++){if(arr[i]==arr[j]){flag=1;}}if(arr[i]<0){//避免负数flag=1;}}//没有重复的数字进入且符合倒数条件if(flag==0&&(a*b*c*(e*f+f*d+d*e)==d*e*f*(b*c+c*a+a*b))){sum++;printf("s=%d;a=%d,b=%d,c=%d,d=%d,e=%d,f=%d\n",s,a,b,c,d,e,f);}}}}}}}printf("sum=%d",sum);}（2）优化后解法#include <stdio.h>void main(){int s;int a,b,c,d,e,f;int arr[7];int flag,i,j;int sum=0;for (s=11;s<=100;s++){printf("s=%d:\n",s);for(a=1;a<=s-2;a++){for(b=a+1;b<=s-1;b++){for(d=a+1;d<=s-2;d++){for(e=d+1;e<=s-1;e++){c=s-a-b;f=s-d-e;//检查是否有重复数字arr[1]=a;arr[2]=b;arr[3]=c;arr[4]=d;arr[5]=e;arr[6]=f;flag=0;for(i=1;i<7;i++){for(j=i+1;j<7;j++){if(arr[i]==arr[j]){flag=1;}}if(arr[i]<0){//避免负数flag=1;}}if(a>b||a>c||b>c){flag=1;}if(d>e||d>f||e>f){flag=1;}//没有重复的数字进入且符合倒数条件if(flag==0&&(a*b*c*(e*f+f*d+d*e)==d*e*f*(b*c+c*a+a*b))){sum++;printf("s=%d;a=%d,b=%d,c=%d,d=%d,e=%d,f=%d\n",s,a,b,c,d,e,f);}}}}}}}7.八皇后问题（1）穷举法#include <stdio.h>#include <stdlib.h>#include <math.h>#define N 8 //定义棋盘大小void main(void){int numoftimes = 0;int count = 0;int a[N + 1];int flag;int i1, i2, i3, i4, i5, i6, i7, i8, x, y, p;printf("%d皇后问题的穷举法解决",N);for (i1 = 1; i1 <= N; i1++){a[1] = i1;for (i2 = 1; i2 <= N; i2++){a[2] = i2;for (i3 = 1; i3 <= N; i3++){a[3] = i3;for (i4 = 1; i4 <= N; i4++){a[4] = i4;for (i5 = 1; i5 <= N; i5++){a[5] = i5;for (i6 = 1; i6 <= N; i6++){a[6] = i6;for (i7 = 1; i7 <= N; i7++){a[7] = i7;for (i8 = 1; i8 <= N; i8++){a[8] = i8;//检测是否冲突flag = 0;for (x = 1; x < N; x++){for (y = x + 1; y <= N; y++){if (a[x] == a[y] || abs(a[x] - a[y]) == y - x){flag = 1;break;}numoftimes++;}if(flag==1){continue;}}if (flag==0){count++;printf("第%d种解法：", count); for (p = 1; p <= N; p++){printf("%d", a[p]);}printf("\n");}}}}}}}}}printf("共运行计算%d次\n", numoftimes);printf("共找到%d种方案\n", count);system("pause");}（2）递归法#include "stdio.h"#define N 8 //定义棋盘大小static int count,a[N];// count记录当前已找到解的个数// a[N]记录皇后的位置,表示第i行的皇后放在棋盘的第a[i]列int numoftimes = 0;void recursive(int t){//尝试第t行的放置方案int i;int j;int sign = 1;if (t == N + 1){//t == N + 1//表示放下最后一个皇后，找到解法count++;/* 第1种显示方法 */// t == N + 1// 已经得到一个解决方案printf("第%d种解法:", count);//Literal1.Text += string.Format("第{0}种解法:", count); for (i = 1; i <= N; i++){printf("%d",a[i]);//Literal1.Text += string.Format("{0}", a[i]);}printf("\n");//Literal1.Text += "<br/>";/* 第2种显示方法printf("第%d种解法:\n", count);for (i = 0; i < N; i ++){for (j = 0; j < N; j ++){if (j == a[i]){printf("@ ");}else{printf("* ");}}printf("\n");}printf("\n");*/}else{//还没有放下所有的皇后for (i = 1; i <= N; i++){//依次尝试在第t行的第1列到第N列放置a[t] = i;sign = 1;for (j = 1; j <= t - 1; j++){//测试皇后在第t行第a[t]列时是否与前面已放置好的皇后相攻击。