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1.16
void print_descending(int x,int y,int z)//按从大到小顺序输出三个数
{
scanf("%d,%d,%d",&x,&y,&z);
if(x
if(y
if(x
printf("%d %d %d",x,y,z);
}//print_descending
1.17
Status fib(int k,int m,int &f)//求k阶斐波那契序列的第m项的值f
{
int tempd;
if(k<2||m<0) return ERROR;
if(m else if (m==k-1) f=1; else { for(i=0;i<=k-2;i++) temp=0; temp[k-1]=1; //初始化 for(i=k;i<=m;i++) //求出序列第k至第m个元素的值 { sum=0; for(j=i-k;j temp=sum; } f=temp[m]; } return OK; }//fib 分析:通过保存已经计算出来的结果,此方法的时间复杂度仅为O(m^2).如果采用递归编程(大多数人都会首先想到递归方法),则时间复杂度将高达O(k^m). 1.18 typedef struct{ char *sport; enum{male,female} gender; char schoolname; //校名为'A','B','C','D'或'E' char *result; int score; } resulttype; typedef struct{ int malescore; int femalescore; int totalscore; } scoretype; void summary(resulttype result[ ])//求各校的男女总分和团体总分,假设结果已经储存在result[ ]数组中 { scoretype score; i=0; while(result.sport!=NULL) { switch(result.schoolname) { case 'A': score[ 0 ].totalscore+=result.score; if(result.gender==0) score[ 0 ].malescore+=result.score; else score[ 0 ].femalescore+=result.score; break; case 'B': score.totalscore+=result.score; if(result.gender==0) score.malescore+=result.score; else score.femalescore+=result.score; break; ……?……?…… } i++； } for(i=0;i<5;i++) { printf("School %d:\n",i); printf("Total score of male:%d\n",score.malescore); printf("Total score of female:%d\n",score.femalescore); printf("Total score of all:%d\n\n",score.totalscore); } }//summary 1.19 Status algo119(int a[ARRSIZE])//求i!*2^i序列的值且不超过maxint { last=1; for(i=1;i<=ARRSIZE;i++) { a[i-1]=last*2*i; if((a[i-1]/last)!=(2*i)) reurn OVERFLOW; last=a[i-1]; return OK; } }//algo119 分析:当某一项的结果超过了maxint时,它除以前面一项的商会发生异常. 1.20 void polyvalue() { float ad; float *p=a; printf("Input number of terms:"); scanf("%d",&n); printf("Input the %d coefficients from a0 to a%d:\n",n,n); for(i=0;i<=n;i++) scanf("%f",p++); printf("Input value of x:"); scanf("%f",&x); p=a;xp=1;sum=0; //xp用于存放x的i次方 for(i=0;i<=n;i++) { sum+=xp*(*p++); xp*=x;
