Jobdu-1534-数组中第K小的数字[解题代码] | Acm之家


九度-1534-数组中第K小的数字[解题代码] | Acm之家
给定两个整型数组A和B。我们将A和B中的元素两两相加可以得到数组C。
譬如A为[1,2],B为[3,4].那么由A和B中的元素两两相加得到的数组C为[4,5,5,6]。
现在给你数组A和B,求由A和B两两相加得到的数组C中,第K小的数字。

long mid=low+((high-low)>>1);  
//统计小于等于 mid 的个数
11lld cntMin(lld mid){
12    lld cnt = 0;
13    lld j = n-1;
14    for(int i=0; i<m; i++){
15        //由于已排序,可直接利用上一次结果
16        while(j >= 0 && arr1[i] + arr2[j] > mid) j--;
17        cnt += j+1;
18    }
19    return cnt;
20}
21
22lld low,high,mid;
23int main() {
24    while(scanf("%lld%lld%lld", &m,&n,&k) != EOF){
25        for(lld i=0; i<m; i++) scanf("%lld", &arr1[i]);
26        for(lld j=0; j<n; j++) scanf("%lld", &arr2[j]);
27        sort(arr1, arr1+m);
28        sort(arr2, arr2+n);
29         low  = arr1[0] + arr2[0];
30         high = arr1[m-1] + arr2[n-1];
31         lld ans;
32         //二分逼近
33        while(low <= high){
34             mid = (low+high)/2;
35            lld cnt = cntMin(mid);
36            if(cnt >= k){
37                ans = mid; //mid有可能是解
38                high = mid-1;
39            }else
40                low = mid+1;
41        }
42        printf("%lld\n",ans);
43
44    }
45    return 0;
46}
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