#include?<stdio.h>
#define?N?10
int?totalscore(int?*score);
int?maxscore(int?*score);
int?minscore(int?*score);
int?averagescore(int?*score);
void?sort(int?*a,?int?left,?int?right);


int?main()
{
????int?score[N]={67,98,75,63,82,79,81,91,66,84};
????printf("totalscore:%d\n",totalscore(score));
????printf("maxscore:%d\n",maxscore(score));
????printf("minscore:%d\n",minscore(score));
????printf("averagescore:%d\n",averagescore(score));
????sort(score,0,9);
????int?i;
????for?(?i?=?0;?i?<?10;?++i)
????{
????????printf("%d?",score[i]);
????????/*?code?*/
????}
????return?0;
}


int?totalscore(int?*score)
{
????int?ts=0,i=0;
????for(i=0;i<10;++i)
????{
????????ts=ts+score[i];
????}
????return?ts;
????
}

int?maxscore(int?*score)
{
????int?max;
????max=score[0];
????int?i=1;
????for(i=1;i<10;++i)
????{
????????if?(score[i]>max)
????????max=score[i];
????}
????return?max;
}

int?minscore(int?*score)
{
????int?min;
????min=score[0];
????int?p=sizeof(score);
????while(p>0)
????{
????????p--;
????????if?(score[p]?<?min)
????????min=score[p];
????}
????return?min;
}
int?averagescore(int?*score)
{
????int?ts=0,i=0;
????for(i=0;i<10;++i)
????{
????????ts=ts+score[i];
????}
????return?ts/10;
}

void?sort(int?*a,?int?left,?int?right)
{
????if(left?>=?right)/*如果左邊索引大于或者等于右邊的索引就代表已經(jīng)整理完成一個組了*/
????{
????????return?;
????}
????int?i?=?left;
????int?j?=?right;
????int?key?=?a[left];
?????
????while(i?<?j)???????????????????????????????/*控制在當(dāng)組內(nèi)尋找一遍*/
????{
????????while(i?<?j?&&?key?>=?a[j])
????????/*而尋找結(jié)束的條件就是，1，找到一個小于或者大于key的數(shù)（大于或小于取決于你想升
????????序還是降序）2，沒有符合條件1的，并且i與j的大小沒有反轉(zhuǎn)*/?
????????{
????????????j--;/*向前尋找*/
????????}
?????????
????????a[i]?=?a[j];
????????/*找到一個這樣的數(shù)后就把它賦給前面的被拿走的i的值（如果第一次循環(huán)且key是
????????a[left]，那么就是給key）*/
?????????
????????while(i?<?j?&&?key?<=?a[i])
????????/*這是i在當(dāng)組內(nèi)向前尋找，同上，不過注意與key的大小關(guān)系停止循環(huán)和上面相反，
????????因為排序思想是把數(shù)往兩邊扔，所以左右兩邊的數(shù)大小與key的關(guān)系相反*/
????????{
????????????i++;
????????}
?????????
????????a[j]?=?a[i];
????}
?????
????a[i]?=?key;/*當(dāng)在當(dāng)組內(nèi)找完一遍以后就把中間數(shù)key回歸*/
????sort(a,?left,?i?-?1);/*最后用同樣的方式對分出來的左邊的小組進(jìn)行同上的做法*/
????sort(a,?i?+?1,?right);/*用同樣的方式對分出來的右邊的小組進(jìn)行同上的做法*/
???????????????????????/*當(dāng)然最后可能會出現(xiàn)很多分左右，直到每一組的i?=?j?為止*/
}

你把你的代碼完整的貼一下，你給的代碼太少，看不明白哦