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改变循环执行的状态,循环程序举例
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?1.??用?break?語(yǔ)句提前終止循環(huán) ?#include?"stdafx.h" ?#define?SUM?100000??????????????????????//指定符號(hào)常量SUM代表100000 ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????float?amount,aver,total; ?????int?i; ?????for(i=1,total=0;i<=1000;i++) ?????{ ?????????printf("please?enter?amount:"); ?????????scanf_s("%f",&amount); ?????????total=total+amount; ?????????if(total>=SUM)break; ?????} ?????aver=total/i; ?????printf("num=%d\naver=%10.2f\n",i,aver); ?????return?0; ?} ???break?其作用是使流程跳到循環(huán)體之外,接著執(zhí)行循環(huán)體下面的語(yǔ)句。 ??2.??用continue語(yǔ)句提前結(jié)束本次循環(huán) ?有時(shí)并不希望終止整個(gè)循環(huán)的操作,而只是希望提前結(jié)束本次循環(huán),而接著 ?執(zhí)行下次循環(huán)。這時(shí)可以用continue語(yǔ)句。 ?#include?"stdafx.h" ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????int?n; ?????for?(n=100;n<=200;n++) ?????{if(n%3==0) ?????continue; ?????printf("%d",n); ?????} ?????printf("\n"); ?????return?0; ?} ???continue; ?其作用為結(jié)束本次循環(huán),即跳過(guò)循環(huán)體中下面尚未執(zhí)行的語(yǔ)句,轉(zhuǎn)到循環(huán)體結(jié)束點(diǎn)之前, ?接著執(zhí)行for語(yǔ)句中的"表達(dá)式3",然后進(jìn)行下一次是否執(zhí)行循環(huán)的判定。 ??3.??break?語(yǔ)句和?continue?語(yǔ)句的區(qū)別 ?continue?語(yǔ)句只結(jié)束本次循環(huán),而不是終止整個(gè)循環(huán)的執(zhí)行。 ?break?語(yǔ)句則是結(jié)束整個(gè)循環(huán)過(guò)程,不再判斷執(zhí)行循環(huán)的條件是否成立。 ??如果有以下兩個(gè)循環(huán)結(jié)構(gòu): ?a.while(表達(dá)式1) ?{ ?????... ?????????if(表達(dá)式2)break; ?????... ?} ??b.while(表達(dá)式1) ?{ ?????... ?????????if(表達(dá)式2)continue; ?????... ?} ??輸出以下?4*5?的矩陣 ?#include?"stdafx.h" ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????int?i,j,n=0; ?????for?(i=1;i<=4;i++) ?????????for(j=1;j<=5;j++,n++) ?????????????{if(n%5==0)printf("\n"); ?????printf("%d\t",i*j); ?????} ?????printf("\n"); ?????return?0; ?} ??#include?"stdafx.h" ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????int?i,j,n=0; ?????for?(i=1;i<=4;i++) ?????????for(j=1;j<=5;j++,n++) ?????????????{if(n%5==0)printf("\n"); ?????if(i==3&&j==1)break; ?????printf("%d\t",i*j); ?????} ?????printf("\n"); ?????return?0; ?} ??#include?"stdafx.h" ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????int?i,j,n=0; ?????for?(i=1;i<=4;i++) ?????????for(j=1;j<=5;j++,n++) ?????????????{if(n%5==0)printf("\n"); ?????if(i==3&&j==1)continue; ?????printf("%d\t",i*j); ?????} ?????printf("\n"); ?????return?0; ?} ???#include?"stdafx.h" ?#include?"math.h" ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????int?sign=1;??????????????????????????double?pi=0.0,n=1.0,term=1.0;?????????????????????????????????????????????while(fabs(term)>=1e-6)??????????????{ ?????????pi=pi+term;??????????????????????????n=n+2;???????????????????????????????sign=-sign;??????????????????????????term=sign/n;?????????????????????} ?????pi=pi*4;?????????????????????????????printf("pi=%10.8f\n",pi);????????????return?0; ?} ??#include?"stdafx.h" ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????int?f1=1,f2=1,f3; ?????int?i; ?????printf("%12d\n%12d\n",f1,f2); ?????for(i=1;i<=38;i++) ?????{ ?????????f3=f1+f2; ?????????printf("%12d\n",f3); ?????????f1=f2; ?????????f2=f3; ?????} ?????return?0; ?} ??#include?"stdafx.h" ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????int?f1=1,f2=1; ?????int?i; ?????for(i=1;i<=20;i++) ?????{ ?????????printf("%12d?%12d",f1,f2); ?????????if(i%2==0)?printf("\n"); ?????????f1=f1+f2; ?????????f2=f2+f1; ?????} ?????return?0; ?} ??輸入一個(gè)大于3的證書(shū)n,判定它是否為素?cái)?shù)(prime,又稱) ?#include?"stdafx.h" ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????int?n,i; ?????printf("please?enter?a?integer?number,n=?"); ?????scanf_s("%d",&n); ?????for(i=2;i<=n-1;i++) ?????????if(n%i==0)break; ?????if(i<n)?printf("%d?is?not?a?prime?number.\n",n); ?????else?printf("%d?is?a?prime?number.\n",n); ?????return?0; ?} ??#include?"stdafx.h" ?#include?"math.h" ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????int?n,i,k; ?????printf("please?enter?a?integer?number:n=?"); ?????scanf_s("%d",&n); ?????k=sqrt((double)n); ?????for(i=2;i<=k;i++) ?????????if(n%i==0)break; ?????if(i<=k)?printf("%d?is?not?a?prime?number.\n",n); ?????else?printf("%d?is?a?prime?number.\n",n); ?????return?0; ?} ??求?100~200?間的全部素?cái)?shù) ?#include?"stdafx.h" ?#include?"math.h" ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????int?n,k,i,m=0; ?????for(n=101;n<=200;n=n+2) ?????{k=sqrt((double)n); ?????for(i=2;i<=k;i++) ?????????if(n%i==0)break; ?????if(i>=k+1) ?????{printf("%d",n); ?????m=m+1; ?????} ?????if(m%10==0)?printf("\n"); ?????} ?????printf("\n"); ?????return?0; ?} ??譯密碼 ?#include?"stdafx.h" ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????char?c; ?????c=getchar();?????????????????while(c!='\n')???????????????{ ?????????if((c>='a'&&c<='z')||(c>='A'&&c<='Z'))??????????????????{ ????????????if(c>='W'&&c<='Z'||c>='w'&&c<='z')?c=c-22;???????????????else?c=c+4;???????????????} ?????printf("%c",c);??????????????c=getchar();?????????????????} ?????printf("\n"); ?????return?0; ?} ??#include?"stdafx.h" ???int?_tmain(int?argc,?_TCHAR*?argv[]) ?{ ?????char?c; ?????while((c=getchar())!='\n')???????????????{ ?????????if((c>='A'&&c<='Z')||(c>='a'&&c<='z'))??????????????????{?c=c+4; ????????????if(c>='Z'&&c<='Z'+4||c>'z') ????????????????c=c-26; ?????????} ?????????printf("%c",c); ?????} ?????printf("\n"); ?????return?0; ?}? ?
轉(zhuǎn)載于:https://blog.51cto.com/290692/1128071
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