算法作業(yè)整理(自己代碼).docx

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1、#include#include#includeusingnamespacestd;intmain(){?intn,i,x[10001],y[10001];?intx_sum,y_sum;?while(cin>>n)?{???for(i=0;i>x[i]>>y[i];????sort(x,x+n);????sort(y,y+n);????for(i=0;i

2、??{??????x_sum+=abs(x[i]-x[n/2]);??????y_sum+=abs(y[i]-y[n/2]);????}????cout<<(x_sum+y_sum)<usingnamespacestd;intmain()?{???//兩個數(shù)組???intlen=0;???//獲取數(shù)組的長度???cin>>len;??//動態(tài)分配數(shù)組???int*a;???int*b;???a=(int*)malloc(sizeof(int)*len);???b=(int*)malloc(sizeof(in

3、t)*len);???//獲取每個數(shù)組的元素???for(inti=0;i>a[i];???}???for(intj=0;j>b[j];??}???//兩個數(shù)組的左右端點(diǎn)的坐標(biāo)???intaLeft=0;???intaRight=len-1;???intbLeft=0;???intbRight=len-1;???//兩數(shù)組中間坐標(biāo)???intaMid=0;???intbMid=0;???//迭代循環(huán)???while(true)???{?????//printf("aleftis%drightis%d,and

4、bleftis%drightis%d.",aLeft,aRight,bLeft,bRight);?????//?如果兩個數(shù)組都只剩下兩個元素，則中位數(shù)一定在其中?????if((aRight-aLeft)==1&&(bRight-bLeft)==1)?????{???????cout<<((a[aLeft]>=b[bLeft])?a[aLeft]:b[bLeft])<<""<<((a[aRight]<=b[bRight])?a[aRight]:b[bRight])<

5、?求解各個數(shù)組的中值???????aMid=(int)((aLeft+aRight)/2);???????bMid=(int)((bLeft+bRight)/2);???????//?如果A中值小于B中值???????if(a[aMid]

6、??????bRight=bMid;?????????}???????}???????//?如果B中值小于A中值???????else???????{?????????//?如果A中現(xiàn)存的數(shù)列是偶數(shù)個，右邊值加一?????????if((aLeft+aRight+1)%2==0){???????????aRight=aMid+1;???????????bLeft=bMid;?????????}?????????else{???????????aRight=aMid;???????????bLeft=bMid;?????????}????????}?????}???}???return0;?}?

7、//中位數(shù)問題O（logn）時間#include#includeusingnamespacestd;intmax(inta,intb){????if(a>=b)????returna;????else??returnb;}intmain(){???????int**a;????int**b;????intn=0;????cin>>n;????a=(int**