Nh p xâu van b n Ð m xem kí t s nào xu t hi n nhi u nh t? Vd: ‘toi di cho t 35 n mua 2 533 d xoi, 2 3 55 d thit. Con 55 0 nghin‘ Kq : ‘ 5 ’ xh nhi u nh t = 6 l n 1/Kb, Nh p xâu 2/Ð m s l n xu t hi n c a m i kí t s (l uu vào 10 bi n l uu k t qu ) *Dùng m ng 10 ph n t (ng m quy u c pt ch s 0 d m s l n xu t hi n c a kí t ‘0’ Ch s i l uu sl xh kt i) *Ð m: xem xét t ng kí t (while) N u l à kt s : tang bi n d m t uong ng lên 1 don v Kí t s[i] l à kt s Tang ph n t a[] l ên 1 don v s[i]==’0’ tang a[0 ] lên 1dv s[i]==’1’ tang a[1] lên 1dv a[(int)s[i] - (int)’0’]++ (int)s[i] - (int)’0’ ==0 n u s[i]==’0’ 1 ‘1’ 2 ‘9’ 3/tìm max trong 10 kq dó =>max In k t qu #include <iostream.h> void main(){ char s[100]; cout<<"Nhap xau: "; cin .getline(s,100); int a[10],i; for(i=0;i<10;i++)a[i]=0; i=0; while(s[i]!='\ 0'){ if(s[i]>='0' && s[i]<='9') a[(int)s[i]-(int)'0']++; i++; } int t; t=0; //t la chi so phat tu max for(i=1;i<10;i++) if(a[i]>a[t])t=i; cout<<"ki tu "<<t<<" xh nhieu nhat voi so lan="<<a[t]; } ./// in ra nhieu kq trung nhau #include <iostream.h> void main(){ char s[100]; cout<<"Nhap xau: "; cin.getline(s,100); int a[10],i; for(i=0;i<10;i++)a[i]=0;//reset a ve 0 i=0; while(s[i]!='\ 0'){//duyet den het xau if(s[i]>='0' && s[i]<='9')//s[i] la kt so a[(int)s[i]-(int)'0']++; i++; } int t; t=0; //t la chi so phan tu max for(i=1;i<10;i++) if(a[i]>a[t])t=i; for(i=0;i<10;i++) if(a[i]==a[t]) cout<<"ki tu "<<i<<" "; cout<<" xh nhieu nhat voi sl="<<a[t]; } . a[1] lên 1dv a[(int)s[i] - (int)’0’]++ (int)s[i] - (int)’0’ ==0 n u s[i]==’0’ 1 ‘1’ 2 9 3/tìm max trong 10 kq dó =>max In k t qu #include <iostream.h> void main(){. for(i=0;i<10;i++)a[i]=0; i=0; while(s[i]!=' 0'){ if(s[i]>='0' && s[i]<=&apos ;9& apos;) a[(int)s[i]-(int)'0']++; i++; } int t; t=0; //t la chi so phat tu max. while(s[i]!=' 0'){//duyet den het xau if(s[i]>='0' && s[i]<=&apos ;9& apos;)//s[i] la kt so a[(int)s[i]-(int)'0']++; i++; } int t; t=0; //t la chi