美利達(dá)達(dá)道路
第一題：#include<>
void?main()
{
int?a[10],b[9],i=0;
for(;i<10;i++)
{
printf("請輸入第%d個數(shù)",i+1);
scanf("%d",&a[i]);
}
for(i=1;i<10;i++)
b[i-1]=a[i]-a[i-1];
for(i=0;i<9;i++)
{
if(i%3==0)printf("\n");
printf("%d? ",b[i]);
}
運行結(jié)果：
第二題：#include<>
void?main()
{
char?a[100];int?i=0;
printf("請輸入一串字符按回車結(jié)束");
scanf("%s",a);
while(1)
{
if(a[i]>='A'&&a[i]<='Z')
a[i]=a[i]+32;
printf("%c",a[i]);
i++;
if(a[i]==0)break;
}
}
運行結(jié)果：
第三題：（以兩名學(xué)生為例，其他只要改變對應(yīng)的值即可，照著框架來即可）#include<>
void?main()
{
int?a[2][5],i,j;
for(i=0;i<2;i++)
{
for(j=0;j<5;j++)
{
printf("請輸入第%d個學(xué)生，第%d門成績",i+1,j+1);
scanf("%d",&a[i][j]);
}
printf("\n");
}
int?sum,csum=0;
printf("%14c第1門課%4c第2門課%4c第3門課%4c第4門課%4c第5門課%4c總分\n",32,32,32,32,32,32);
for(i=0;i<2;i++)
{
sum=0;
printf("第%d個學(xué)生",i);
for(j=0;j<5;j++)
{
printf("%6c%d",32,a[i][j]);
printf("%3c",32);
sum=sum+a[i][j];
}
csum=csum+sum;
printf("%5c%d",32,sum);
printf("\n");
}
int?temp;
printf("班級總分?");
for(j=0;j<5;j++)
{
temp=0;
for(i=0;i<2;i++)
temp=temp+a[i][j];
printf("%6c%d",32,temp);
printf("%2c",32);
}
printf("%5c%d\n",32,csum);
printf("\n");
運行結(jié)果：
第四題：#include<>
void?main()
{
int?a[10],i=0,j,temp;
for(;i<10;i++)
{
printf("請輸入第%d個數(shù)",i+1);
scanf("%d",&a[i]);
}
for(j=0;j<=9;j++)
{
for?(i=0;i<10-j;i++)
if?(a[i]>a[i+1])
{
temp=a[i];
a[i]=a[i+1];
a[i+1]=temp;
}
}
for(i=0;i<10;i++)
printf("%d?",a[i]?);
printf("\n");
運行結(jié)果：
學(xué)習(xí)是要用心的，多動腦，多思考學(xué)到的東西才是自己的。喔，對了給分喲?。。。?！
豬寶0517
1、#include <>main(){ int a, b, square; scanf (“%d%d”, &a, &b); square = a * a + b * b; if (square > 100) \*判斷a、b的平方和是否大于100 *、 { printf (“their square is bigger than 100\n”); printf (“the digitale bigger than 100 is :%d”, square / 100); } else { printf (“their square is smaller than 100\n”); printf (“their addtion is: %d”, a + b); }}2、#include <>main(){ int n; if ((n % 5 == 0) && (n % 7 == 0)) \* 判斷n是否為5和7的公倍數(shù) *\ { printf (“5 and 7 yes”); } else { if (n % 3 == 0) \* 判斷是否能被3整除 *\ printf (“ 3 yes”); else printf (“no”); }}3、(3題和平共處題與第2題相似的，只要把條件改一下就可以了）#include <>main(){ int n; if ((n % 3 == 0) && (n % 5 == 0)) { printf (“3 and 5 yes”); } else { if (n % 7 == 0) printf (“ 7 yes”); else printf (“no”); }}4、#include <>main(){ int n; if ((n % 2 == 0) && (n % 3 == 0)) { printf (“2&3 yes”); } else { if (n % 7 == 0) printf (“ 3 yes”); else printf (“no”); }}5、#include <>main(){ int x, y; printf ("x ="); scanf ("%d", &x); if (x == 2) \* 用一個多分支語句將幾種情況分開計算*\ y = 2 * x; else if (x < 2) y = x * x + 1; else y = 2 * x * x + 3 * x + 1; printf ("y = %d", y);}6、（6 題和7題還有8題都與5題相似）#include <>main(){ int x, y; printf ("x ="); scanf ("%d", &x); if (x == 1) y = 1; else if (x < 1) y = x * x; else y = x * x * x; printf ("y = %d", y);}7、#include <>main(){ int x, y; printf ("x ="); scanf ("%d", &x); if (x < 0) y = x + 1; else if (x <= 2) y = x * x + 2; else y = x * x * x + 3; printf ("y = %d", y);}8、#include <>main(){ int x, y; printf ("x ="); scanf ("%d", &x); if (x < 1) y = x; else if (x <= 10) y =2 * x - 1; else y = 3 * x - 11; printf ("y = %d", y);}9、#include <>main(){ float score; char grade; printf ("please enter the score:"); scanf ("%f", &score); if (score > 100 || score < 0) \*判斷成績是否輸入正確*\ { printf ("enter error!"); } if (score >= 90) grade = 'A'; else if (score >= 80) \*在這里的else if 語句條件中已經(jīng)排除了在于等于90的情況，即此處等同于"score >= 80 && score < 90"*\ grade = 'B'; else if (score >= 70) grade = 'C'; else if (score >= 60) grade = 'D'; else grade = 'E'; printf ("the grade is :%c", grade);}10、#include <>main(){ int n, price; printf ("please enter quantity:"); scanf ("%d", &n); if (n <= 10) price = 60; else if (n < 40) \* 此處方法與上題相同*\ price = 50; else price = 45; printf ("the total money is :%d", price * n);}11、#include <>#include <>main(){ int a, b,final; printf ("enter a,b:"); scanf ("%d%d", &a, &b); if (a % b == 0) final = a * a + b * b; else if ( b % a == 0) final = a * a * a + b * b * b; else final = abs(a - b); printf ("final = %d", final);}12、#include <>#include <>main(){ float a, b, f; printf ("enter a,b:"); scanf ("%f%f", &a, &b); if (a > b) f = fabs(a - b); else if (a < b) f = a * b; else f = ((int)a % 10) * ((int)b % 10); \*用強(qiáng)制轉(zhuǎn)換將a、b轉(zhuǎn)換成整數(shù)再除10求余即得個位數(shù)字*\ printf ("f = %.2f", f); }13、 #include <>#include <>main(){ int a, b, c, disc; int x1, x2，x; printf ("a,b,c="); scanf ("%d%d%d", &a, &b, &c); if (a == 0) { printf ("this is not a equation"); } else { disc = b * b - 4 * a * c; if ( disc > 0) { x1 = （- b + sqrt(disc)) / （2 * a); x2 = - b - sqrt(disc) / (2 * a); printf ("there are two deferent root:%d %d", x1, x2）； } else { x = - b / ( 2 * a); printf ("there are two same root:%d", x); } }}14、#include <>main(){ int n; int n1, n2, n3; printf ("please enter a number:"); scanf ("%d", &n); if (n >= 100 && n <= 999) { n1 = n / 10;\*取出n的個、十、百各位數(shù)*\ n2 = (n - n1* 10) / 10； n3 = n % 10; if (n1 * n1 * n1 + n2 * n2 * n2 + n3 * n3 * n3 == n)\* 判斷個、十、百位平方和是否等于n *\ printf ("the number is shuixianhua digitale"); else printf ("the number is not shuixianhua digitale"); } else printf ("the number is not a 3 bit number");}15、#include <>main(){ enum workday {monday = 1, tuesday, wednesday, thursday, friday, saturday, sunday}; enum workday workdays; int n; printf ("please enter a days:"); scanf ("%1d", &n); if (n >= 1 && n <= 7) { workdays = (enum workday)n; switch(workdays) { case 1: printf ("moday"); break; case 2: printf ("tuesday"); break; case 3: printf ("wednesday"); break; case 4: printf ("thursday"); break; case 5: printf ("friday"); break; case 6: printf ("saturday"); break; case 7: printf ("sunday"); break; } } else printf ("enter error");} 16、#include <>main(){ int i, n; float t, s = 0; \* 用t 產(chǎn)生各項，s 為各項之和 *\ printf ("n = "); scanf ("%d", &n); for (i = 1; i <= n; i++) { t = / n; \* 因為n 是整數(shù)，在1后面加小數(shù)使得結(jié)果不至于為零*\ s = s + t; n = n + 2; } printf ("the addtion is :%f", s);}
曼特寧先森
#include <>
#include <>
#include <>
int main()
{
int n,*a,i,j,t;
srand(time(NULL));
do
{
scanf("%d",&n);
}while(n<1);
a=(int*)malloc(sizeof(int)*n);
for(i=0;i a[i]=rand()%99+1; for(i=0;i printf("%d%c",a[i],(i+1)%10?' ':'\n'); printf("\n"); for(i=0,j=n-1;i { t=a[i]; a[i]=a[j]; a[j]=t; } for(i=0;i printf("%d%c",a[i],(i+1)%10?' ':'\n'); printf("\n"); free(a); return 0; }
dp73732849
1輸入兩個整數(shù)a和b，若a和b的平方和大于100，則輸出平方和的百位以上的數(shù)字，否則輸出a和 b的和。 要求： 1）輸出結(jié)果時說明平方和是大于100還是小于100（ >100或<100 ）#include<>int main(）{ int a, b, c; scanf("%d%d", &a, &b); c = a * a + b * b; if(c > 100) printf("平方和大于100\n%d\n", c%100); else printf("平方和小于等于100\n%d\n", a + b); return 0;} 2輸入一個整數(shù)，判斷是否是5和7的公倍數(shù)，若是則輸出：5and7yes，否則再判斷是否是3的倍數(shù)，若是3的倍數(shù)輸出：3yes，若都不是則輸出：no#include<>int main(){ int input; scanf("%d", &input); if(input % 5 == 0 && input % 7 == 0)printf("5and7yes\n"); else if(input % 3 == 0) printf("3yes\n"); else printf("no\n"); return 0;}第三，四題和第二題差不多 5計算公式： [ 2*x x=2 y= [ x*x+1 x<2 [ 2*x*x+3*x+1 x>2 要求： 1）從鍵盤輸入x的值，根據(jù)x的值求y的值 2）輸出y的值#include<>int main(){ int x, y; scanf("%d", &x); if(x == 2) y = 2*x; else if(x < 2) y = x*x+1; else y = 2 * x * x + 3 * x + 1; printf("%d\n", y); return 0;}8 和7差不多 下面原理都差不多 用if基本都能搞定 判斷條件就行了 應(yīng)該LZ能搞定了 東西有點多 -。- 就寫這些吧 其他的就不一一寫了^ ^
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