if (k<>j) {x=a[i];a[i]=a[k];a[k]=x;}
}
}/* sort */
O(n2)
(5)void matrimult(a[m][n],b[n][l],c[m][l],int m,int n,int l){ /* a为m×n阶的矩阵,b为n×l阶的矩阵,c为m×l阶的矩阵*/ /* 计算c=a*b */
for (i=1; i<=m; ++i)
for (j=1; j<=l; ++j)
c[i,j]=0;
for (i=1; i<=m; ++i)
for (j=1; j<=l; ++j)
for (k=1; k<=n; ++k)
c[i,j]=c[i,j]+a[i,k]*b[k,i];
}/* matrimult */
O(n3)
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(xy; //<->为表示交换的双目运算符,以下同
if(yz;
if(xy; //冒泡排序
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(melse if (m==k-1) f=1;
else
{
for(i=0;i<=k-2;i++) temp[i]=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[i]=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(m2).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[i].sport!=NULL)
{
switch(result[i].schoolname)
{
case 'A':
score[ 0 ].totalscore+=result[i].score;
if(result[i].gender==0) score[ 0 ].malescore+=result[i].score;
else score[ 0 ].femalescore+=result[i].score;
break;
case 'B':
score.totalscore+=result[i].score;
if(result[i].gender==0) score.malescore+=result[i].score;
else score.femalescore+=result[i].score;
break;
}
i++;
}
for(i=0;i<5;i++)
{
printf("School %d:\n",i);
printf("Total score of male:%d\n",score[i].malescore);
printf("Total score of female:%d\n",score[i].femalescore);
printf("Total score of all:%d\n\n",score[i].totalscore);
}
}//summary
1.19
Status algo119(int a[ARRSIZE])//求i!*2^i序列的值且不超过maxint
{
last=1;