#include <stdio.h>
#include <math.h>
void deal(int a,int b,int c){if(a==0&&b==0&&c==0)printf("a=%d,b=%d,c=%d时方程为无穷解\n",a,b,c);else if(a==0&&b==0&&c!=0)printf("a=%d,b=%d,c=%d时方程无解\n",a,b,c);elseprintf("a=%d,b=%d,c=%d时方程无解\n",a,b,c);
}
void deal(int a,int b,int c,double x1){printf("a=%d,b=%d,c=%d时方程为单解x=%lf\n",a,b,c,x1);
}
void deal(int a,int b,int c,double x1,double x2){printf("a=%d,b=%d,c=%d时方程解为x1=%lf,x2=%lf\n",a,b,c,x1,x2);
}
int main(){int a,b,c;printf("please input the a,b,c:\n");scanf("%d%d%d",&a,&b,&c);if(a==0&&b==0&&c==0){deal(a,b,c);}else if(a==0&&b==0&&c!=0){deal(a,b,c);}else if(a==0&&b!=0&&c==0){deal(a,b,c,0);}else if(a!=0&&b==0&&c==0){deal(a,b,c,0);}else if(a==0&&b!=0&&c!=0){deal(a,b,c,c*1.0/b);}else if(a!=0&&b!=0&&c==0){deal(a,b,c,0,-b*1.0/a);}else if(a!=0&&b==0&&c!=0){double dis=b*b-4.0*a*c;if(dis<0) deal(a,b,c);else deal(a,b,c,sqrt(a*1.0/c),-sqrt(a*1.0/c));}else if(a!=0&&b!=0&&c!=0){double dis=b*b-4*a*c;if(dis<0) deal(a,b,c);else deal(a,b,c,(-b-sqrt(dis))/(2*a),(-b+sqrt(dis))/(2*a)); } return 0;
}

#include <stdio.h>
#define N 5
int a[N][N];
void Input();//输入数组
int AnDian();//寻找安点
int main(){printf("请输入五阶矩阵:\n");Input();int k=AnDian();if(k==0){printf("NO FIND!\n");}return 0;
}
void Input(){int i,j;for(i=0;i<N;i++){for(j=0;j<N;j++){scanf("%d",&a[i][j]);}}
}
//寻找安点;
int AnDian(){int i,j,k,r,c,value,count=0;for(i=0;i<N;i++){//行 value=a[i][0];r=i;c=0;for(j=0;j<N;j++){//找出每行的最大值 if(value<a[i][j]){value=a[i][j];r=i;c=j;} }for(k=0;k<N;k++){//遍历行最小值的所在列,判断是否依然是列的最小值 if(value>a[k][c]){value=a[k][c];r=k;}}if(r==i){//如果最终行没有变化则说明行的最大值也是列的最小值。 printf("%d(%d,%d)\n",value,r+1,c+1);count++;}}if(count==0) return 0;else return 1;
}

#include <stdio.h>
#include <math.h>
int main(){int fz=2,fm=1,i,j,t;double sum=0.0;for(i=1;i<=20;i++){sum+=1.0*fz/fm;t=fz;fz=fz+fm;fm=t;}printf("%lf\n",sum);return 0;
} 

#include <stdio.h>
#include <math.h>
#define M 30
struct student{int no;//学号char name[20];//姓名double score[5];//成绩
}stu[M];
void OneAvg();
void GradeL();
void AvgMore();
int main(){int i,j;printf("请输入30个学生的信息\n");for(i=0;i<M;i++){printf("请输入第%d个学生的信息:\n",i+1);scanf("%d%s",&stu[i].no,stu[i].name);for(j=0;j<5;j++){scanf("%lf",&stu[i].score[j]);}}OneAvg(); GradeL();AvgMore();return 0;
}
void OneAvg(){int i,j;double sum=0;for(i=0;i<M;i++)sum+=stu[i].score[0];printf("第一门课的平均分是%lf\n",sum/M);
}
void GradeL(){//有两门课不及格的学生的信息 int i,j;printf("有两门课成绩不合格的学生信息如下:\n");printf("学号\t姓名\t课程1\t课程2\t课程3\t课程4\t课程5\n");for(i=0;i<M;i++){int count=0;for(j=0;j<5;j++){if(stu[i].score[j]<60) count++;    }if(count>2){printf("%d\t%s\t",stu[i].no,stu[i].name);for(j=0;j<5;j++){printf("%lf\t",stu[i].score[j]);}printf("\n");}}
}
void AvgMore(){//平均分超90或者所有的课程成绩都在85以上 int i,j; printf("平均分大于90或者课程成绩都在85以上的学生:\n");printf("学号\t姓名\t课程1\t课程2\t课程3\t课程4\t课程5\n");for(i=0;i<M;i++){double sum=0;int count=0;for(j=0;j<5;j++){sum+=stu[i].score[j];if(stu[i].score[j]>85) count++;}if((sum/5>90)||(count==5)){printf("%d\t%s\t",stu[i].no,stu[i].name);for(j=0;j<5;j++){printf("%lf\t",stu[i].score[j]);}printf("\n");}}
} 

2001 liu 85 85 85 76 100
2002 bao 90 90 90 90 90
2003 wang 100 100 100 100 70
2004 chen 70 100 80 90 60
2005 hu 100 50 100 90 100
2006 xu 70 90 90 90 90
2007 hua 10 10 10 10 10
2008 ma 20 20 20 20 20
2009 luo 30 30 30 30 30
2010 l1 40 40 40 40 40
2011 l2 50 50 50 50 50
2012 l3 60 60 60 60 60
2013 l4 70 70 70 70 70
2014 l5 80 80 80 80 80
2015 l6 90 90 90 90 90
2016 l7 40 40 40 40 40
2017 l8 50 50 50 50 50
2019 l9 60 60 60 60 60
2020 l10 70 70 70 70 70
2021 l11 80 80 80 80 80
2022 l12 90 90 90 90 90
2023 l13 40 40 40 40 40
2024 l14 50 50 50 50 50
2025 l15 60 60 60 60 60
2026 l41 70 70 70 70 70
2027 l51 80 80 80 80 80
2028 l61 90 90 90 90 90
2029 l52 80 80 80 80 80
2030 l62 90 90 90 90 90
2031 l62 90 90 90 90 90