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Chng 4: Cu lnh lpMc 7: gii bi tp tng hp

Bi tp nng cao:1.Nhp vo 1 s nguyn dng n, m xem n c bao nhiu ch s, tmra ch s ln nht, ch s nh nht=================================Cch lm: ly s lin tc chia cho 10, mi ln chia th tng bin m lnchia cho n khi s l 0 th dng livd: s n = 12341234 / 10 = 123 => dem = 1123 / 10 = 12 => dem = 212 / 10 = 1 => dem = 31 / 10 = 0 => dem = 4

int dem = 0;int temp = n; //to bin ph temp lu gi tr nwhile ( temp != 0){ temp = temp/ 10; // Lin tc chia cho 10a dem++;}printf("\nSo %d co %d chu so", n, dem);

-------------------------------------------Cch 2: //dem = log10((double)n) + 1;if(n == 0){ dem = 1;}else{ dem = log10((double)n) + 1;}/*1234 lm cch no ra c cc ch s 1, 2, 3, 41234 % 10 = 4 (1)1234 / 10 = 123

123 % 10 = 3(2)123 / 10 = 12

12 % 10 = 2 (3)12 / 10 = 1

1 % 10 = 1 ( 4)1 / 10 = 0 =>dng li*/

int max , min;//123 % 10 = 3

max = min = temp % 10; // t max = min = s cuitemp /= 10; // b s cuiwhile(temp != 0){ int chuso = temp; temp /= 10;if( chuso > max) { max = chuso; }

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if ( chuso < min) { min = chuso; }}

v d: 5207B1: max = min = 75207 = 10 = 520

520 % 10 = 0 => min = 0, max = 7520 / 10 = 52

52 % 10 = 2 => min = 0, max = 752 / 10 = 5---------------------------------------------/*Nu khng t max = min = temp % 10 th t max l ch s nh nht m bo mi chkhi ly ra lun > max, v t min l ch s ln nht m bo mi ch s khi ly ra lmax = 0min = 9v ch c 9 ch s t 0 -> 9 => 0 l nh nht, 9 l ln nht

5 % 10 = 5 => min = 0 , max = 75 / 10 = 0 => iu kin dng

TM KIU LN NHT/B NHT CA INT#include

INT_MAX => s ln nht ca kiu intINT_MIN => s b nht ca kiu int-----------------------------------------TM KIU LN / B NHT CA S THCFLT_MAX => kiu floatFLT_MIN => kiu float

----------------------------------------DBL_MAX => kiu doubleDBL_MIN => kiu double

--------------------------------------------------------------------------------------int n;do{ printf("\nNhap vao so nguyen duong n = "); scanf("%d", &n); if(n < 0) { printf("\nGia tri n khong hop le.Xin kiem tra lai !"); }}while( n < 0);--------------------------------------------

- Bi c s:vd:6 => sau16 => "muoi" sau216 => hai "tram" muoi sau3216 => ba "ngan" hai "tram" "muoi" sau=> Nguyn tc ly ra tng ch s mnh ch c nguyn tc l chia % 10 ri / 10 m nh vch ly theo chiu t phi sang, trong khi c s th mnh c t tri sang cho nn trmnh o s li.V d:

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3216 => o li l 6123ri t 6123 c t phi sang tri s c cc ch s l 3, 2, 1, 6====================================================BI TP:1.S i xng: mun chng minh mt s c l i xng hay khng ta tm ra s o ngc cnh s o ngc bng chnh s ban u => n l i xng

abcd => o li l : dcba1234 => o li l: 43214321 = 4 * 1000 + 3 * 100 + 2 * 10 + 1*1 = 4* 10^3 + 3 * 10^2 + 2*10^1 + 1*10^0-----------------------------------------------123456 => 654321=> 654321 = 6*10^5...MotHaiBaBonNamSauBayTamChin

kim tra ng sau s va c nu cn 3 ch s th s xut ra thm ph m "ngn", nu 2 l "trm", nu 1 ch s l "mi", nu 0 ch s th thi.-----------------------------------------------#include #include #include int main(){do{

printf("\nNhap vao n ( n > 0): ");scanf("%d", &n);

if(n

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printf("\nKhong la so doi xung");}getch();return 0;}