题目来源:https://pintia.cn/problem-sets?tab=0
7-1 厘米换算英尺英寸#include7-2 然后是几点int main(){ int cm; scanf("%d",&cm); int foot=cm/30.48; int inch=(cm/30.48-foot)*12; printf("%d %d",foot,inch); return 0; }
#include7-3 逆序的三位数int main(){ int t1,t2; scanf("%d %d",&t1,&t2); int minute=t1/100*60+t1%100+t2; int h=minute/60; int m=minute%60; printf("%d%02d",h,m); return 0; }
#include7-4 BCD解密int main(){ int i; scanf("%d",&i); int a=i/100; int b=i/10%10; int c=i%10; printf("%d",c*100+b*10+a); return 0; }
#include7-5 表格输出int main(){ int x; scanf("%d",&x); int a = x%16 + x/16*10; printf("%dn",a); }
#include7-6 混合类型数据格式化输入int main(){ printf("------------------------------------n"); printf("Province Area(km2) Pop.(10K)n"); printf("------------------------------------n"); printf("Anhui 139600.00 6461.00n"); printf("Beijing 16410.54 1180.70n"); printf("Chongqing 82400.00 3144.23n"); printf("Shanghai 6340.50 1360.26n"); printf("Zhejiang 101800.00 4894.00n"); printf("------------------------------------n"); return 0; }
#include7-7 12-24小时制int main(){ float f1,f2; int i; char ch; scanf("%f %d %c %f",&f1,&i,&ch,&f2); printf("%c %d %.2f %.2f",ch,i,f1,f2); return 0; }
#include7-8 超速判断int main(){ int h,m; scanf("%d:%d",&h,&m); if(h>=0&&h<12) printf("%d:%d AM",h,m); else if(h==12) printf("%d:%d PM",h,m); else if(h==24) printf("%d:%d AM",h%12,m); else printf("%d:%d PM",h%12,m); return 0; }
#include7-9 用天平找小球int main(){ int s; scanf("%d",&s); if(s>60) printf("Speed: %d - Speeding",s); else printf("Speed: %d - OK",s); return 0; }
#include7-10 计算工资int main(){ int a,b,c; scanf("%d %d %d",&a,&b,&c); if(a==b && a!=c) printf("C"); else if(a==c && a!=b) printf("B"); else if(b==c && b!=a) printf("A"); return 0; }
#include7-11 分段计算居民水费int main(){ int year,time,salary_hour=0; float salary; scanf("%d %d",&year,&time); if(year<=5){ salary_hour = 30; }else salary_hour = 50; if(time<=40){ salary=salary_hour*time; }else salary=salary_hour*40 + (time-40)*salary_hour*1.5 printf("%.2f",salary); return 0; }
#include7-12 两个数的简单计算器int main(){ int x; double y; scanf("%d",&x); if(x<=15) y=(double)4*x/3; else y=2.5*x-17.5; printf("%.2f",y); return 0; }
#includeusing namespace std; int main(){ int a,b; char op; cin>>a>>op>>b; switch(op){ case '+':cout< 7-13 日K蜡烛图 #includeusing namespace std; int main(){ double Open,High,Low,Close; cin>>Open>>High>>Low>>Close; if(Close < Open){ cout<<"BW-Solid"; if(Low < Open && Low < Close && High > Open && High > Close){ cout<<" with Lower Shadow and Upper Shadow"; } else if(Low < Open && Low < Close){ cout<<" with Lower Shadow"; }else if(High > Open && High > Close){ cout<<" with Upper Shadow"; } } else if(Close > Open){ cout<<"R-Hollow"; if(Low < Open && Low < Close && High > Open && High > Close){ cout<<" with Lower Shadow and Upper Shadow"; } else if(Low < Open && Low < Close){ cout<<" with Lower Shadow"; }else if(High > Open && High > Close){ cout<<" with Upper Shadow"; } }else{ cout<<"R-Cross"; if(Low < Open && Low < Close && High > Open && High > Close){ cout<<" with Lower Shadow and Upper Shadow"; } else if(Low < Open && Low < Close){ cout<<" with Lower Shadow"; }else if(High > Open && High > Close){ cout<<" with Upper Shadow"; } } return 0; } 重复代码好多
7-14 求整数段和#includeusing namespace std; int main() { int a,b,cnt,sum=0; scanf("%d %d",&a,&b); if(a<=b){//竟然错在这里一直过不去,我吐吐吐了 for(int i=a;i<=b;i++){ printf("%5d",i); cnt++;sum += i; if(cnt%5==0 || i==b) cout< 7-15 计算圆周率 #include7-16 求符合给定条件的整数集using namespace std; int main(){ double sum=1.0;// pi/2 double num=1.0; double t;// 阈值 cin>>t; for(int i=1;num>t;i++){ num *=(double)i/(2*i+1); sum += num; } printf("%.6f",sum*2); return 0; } #include7-17 爬动的蠕虫using namespace std; int main(){ int a,cnt=0; cin>>a;//a,a+1,a+2,a+3 for(int i=a;i<=a+3;i++){ for(int j=a;j<=a+3;j++){ for(int k=a;k<=a+3;k++){ if(i!=j&&j!=k&&k!=i){ printf("%d",i*100+j*10+k); cnt++; if(cnt%6==0){ printf("n"); }else printf(" "); } } } } return 0; } #includeusing namespace std; int main(){ int n,u,d,i,cnt=0; cin>>n>>u>>d; for(i=1;cnt 感觉有点小毛病
7-18 二分法求多项式单根#include#include using namespace std; double a3, a2, a1, a0; double f(double x) { return a3 * pow(x, 3) + a2 * pow(x, 2) + a1 * pow(x, 1) + a0; } bool divide(double& a, double& b) { if (f((a + b) / 2) == 0||(b-a)<0.001) return true; if (f(a) * f(b) <= 0) { if(f(a)==0){ b=0; a*=2; } if(f(b)==0){ a=0; b*=2; } if (f((a + b) / 2) * f(a) > 0) { a = (a + b) / 2; divide(a, b); } if (f((a + b) / 2) * f(b) > 0) { b = (a + b) / 2; divide(a, b); } } } int main() { double a, b; cin >> a3 >> a2 >> a1 >> a0; cin >> a >> b; divide(a, b); printf("%.2fn", (a + b) / 2); return 0; } 大佬们的代码是如此简洁呜呜呜,不像我呜呜呜
7-19 支票面额#includeusing namespace std; int main(){ int n; cin>>n; for(int y=0;y<100;y++){ for(int f=0;f<100;f++){ if(98*f==199*y+n){ printf("%d.%d",y,f); return 0; } } } printf("No Solution"); return 0; } 说实话没有想到用for循环,呜呜呜



