题目描述
给定2个数组a、b以及一个整数x
判断是否存在a[i]+b[j]=x
输入
第一行：两个整数n、m (1<n、m<10000)
第二行：n个整数，为数组a
第三行：m个整数，为数组b
第四行：一个整数k (1<k<20)，表示k次查询
接下来的k行：每行一个整数x
输出
若数组a中存在一个a[i]，数组b中存在一个b[j]，使得a[i]+b[j]=x，则输出Yes
如不存在，则输出No
样例输入
2 3
1 2
3 4 5
4
1
4
5
8
样例输出
No
Yes
Yes
No
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 using namespace std; 5 const int maxn = 1e5+7; 6 int a[maxn], b[maxn]; 7 int query(int *arr, int l, int r, int key){ 8 while(l <= r){ 9 int mid = (l + r) >> 1; 10 if(arr[mid] == key) return mid; 11 if(arr[mid] < key) l = mid + 1; 12 else r = mid - 1; 13 } 14 return -1; 15 } 16 int main() 17 { 18 int n, m, k; 19 scanf("%d%d",&n,&m); 20 for(int i = 0; i < n; i++)scanf("%d",&a[i]); 21 for(int i = 0; i < m; i++)scanf("%d",&b[i]); 22 cin >> k; 23 sort(a, a + n);sort(b, b + m); 24 int q; 25 while(k--){ 26 cin >> q; 27 int flag = 1; 28 for(int i = 0; i < n; i++){ 29 int y = q - a[i]; 30 //cout << y << endl; 31 if(y > 0){ 32 int pos = query(b, 0, m, y); 33 if(pos != -1){ 34 cout << "Yes" << endl; 35 flag = 0; 36 break; 37 } 38 } 39 40 } 41 if(flag == 1){ 42 cout << "No" << endl; 43 } 44 } 45 return 0; 46 }
670 D2
The term of this problem is the same as the previous one, the only exception — increased restrictions.
The first line contains two positive integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 10^9) — the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 10^9), where the i-th number is equal to the number of grams of the i-th ingredient, needed to bake one cookie.
The third line contains the sequence b1, b2, ..., bn (1 ≤ bi ≤ 10^9), where the i-th number is equal to the number of grams of the i-th ingredient, which Apollinaria has.
Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder.
1 1000000000
1
1000000000
2000000000
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
0
3 1
2 1 4
11 3 16
4
4 3
4 3 5 6
11 12 14 20
3
1 #include <iostream> 2 #include <cstdio> 3 #define ll long long 4 using namespace std; 5 const int maxn = 1e5+7; 6 ll a[maxn], b[maxn]; 7 8 int main() 9 { 10 ll n, k; 11 while(~scanf("%lld%lld",&n,&k)){ 12 for(int i = 0; i < n; i++)scanf("%lld",&a[i]); 13 for(int i = 0; i < n; i++)scanf("%lld",&b[i]); 14 ll mid, sum, flag, l = 0, r = 2e9+10; 15 while(l <= r){ 16 mid = (l + r) >> 1; 17 sum = 0;flag = 1; 18 for(int i = 0; i < n; i++){ 19 if(a[i]*mid > b[i]){ 20 sum += a[i]*mid - b[i]; 21 } 22 if(sum > k){ 23 flag = 0; 24 } 25 } 26 //cout << l << ' ' << r << endl; 27 if(flag) l = mid + 1; 28 else r = mid - 1; 29 } 30 printf("%lld ",l - 1); 31 } 32 33 return 0; 34 }
题目描述
一群妖王排成一排站在苦情巨树下，寻找自己的转世恋人。
虽然都是妖王，但按照涂山的规定必须进行标号，标号为1的妖王排在最后面，标号为n的妖王排在最前面。每个妖王只有一个妖力值a[i]表示它们现在的地位。
妖王们是讲究实力的，当然不服比它妖力值低的居然可以排在前面，它们现在想知道在它前面，妖力值比它低，而且离它最远的距离是多少？
输入
输入n表示有n个妖王
第二行输入n个整数，表示每个妖王的妖力值a[i]
n<=2*105
1<=a[i]<=109
输出
对于每个妖王，妖力值比它低，而且离它最远的距离是多少，如果没有输出-1
样例输入
6
5 50 45 7 10 1
样例输出
4 3 2 1 0 -1
求最远距离的最小值，可以用一辅助数组M把最小值弄好，这样就可以进行二分查找了。
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 const int maxn = 2e5+10; 5 6 int a[maxn],M[maxn]; 7 int MIN(int a,int b){ 8 return a>b?b:a; 9 } 10 int main(){ 11 int n; 12 while(~scanf("%d",&n)){ 13 for(int i = 1; i <= n; i++) scanf("%d",&a[i]); 14 M[n] = a[n]; 15 for(int i = n - 1; i > 0; i--) M[i] = MIN(M[i+1],a[i]); 16 for(int i = 1; i <= n; i++){ 17 int ans = -1,l = i + 1,r = n; 18 while(l <= r){ 19 int mid = (l + r) >> 1; 20 if(a[i] > M[mid]){ 21 ans = mid; 22 l = mid + 1; 23 } 24 else r = mid - 1; 25 } 26 if(ans == -1) printf("-1"); 27 else printf("%d",ans - i - 1); 28 if(i == n) printf(" "); 29 else printf(" "); 30 } 31 } 32 return 0; 33 }