D. Professor GukiZ and Two Arrays 题目连接: http://www.codeforces.com/contest/620/problem/D Description Professor GukiZ has two arrays of integers, a and b. Professor wants to make the sum of the elements in the array a sa as close as possible to the sum…

Professor GukiZ and Two Arrays 题意:两个长度在2000的-1e9~1e9的两个序列a,b(无序);要你最多两次交换元素,使得交换元素后两序列和的差值的绝对值最小:输出这个最小的和的差值的绝对值:并且输出交换次数和交换的序号(从1 开始) Input 5 5 4 3 2 1 4 1 1 1 1 Output 1 2 1 1 4 2 策略: 若是只交换一次,直接O(n^2)暴力即可:但是里面可以交换两次..若是分开看..没思路.那就开始时就预处理出同一个序列中任意两个…

A. Professor GukiZ's Robot   Professor GukiZ makes a new robot. The robot are in the point with coordinates (x1, y1) and should go to the point (x2, y2). In a single step the robot can change any of its coordinates (maybe both of them) by one (decrea…

D. Gadgets for dollars and pounds 题目连接: http://www.codeforces.com/contest/609/problem/C Description Nura wants to buy k gadgets. She has only s burles for that. She can buy each gadget for dollars or for pounds. So each gadget is selling only for som…

题目:http://codeforces.com/contest/1156/problem/E 题意:给你1-n  n个数,然后求有多少个区间[l,r] 满足    a[l]+a[r]=max([l,r]) 思路:首先我们去枚举区间肯定不现实,我们只能取把能用的区间去用,我们可以想下每个数当最大值的时候所做的贡献 我们既然要保证这个数为区间里的最大值,我们就要从两边扩展,找到左右边界能扩展在哪里,这里你直接去枚举肯定不行 这里我们使用了线段树二分去枚举左右区间最远能走到哪里,然后很暴力的去枚举短…

D. Gadgets for dollars and pounds time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Nura wants to buy k gadgets. She has only s burles for that. She can buy each gadget for dollars or for po…

#include<bits/stdc++.h>using namespace std;typedef long long ll;int n,k;ll a[200007],b[200007],s[200007];ll ans;int check(ll val){ memset(s,0,sizeof(s)); int num=k; for(int i=1;i<=n;i++){  if(!b[i])   continue;  s[min(a[i]/b[i]+1,1ll*(k+1))]++;//…

传送门 题意 将n个数划分为两块,最多改变一个数的位置, 问能否使两块和相等 分析 因为我们最多只能移动一个数x,那么要么将该数往前移动,要么往后移动,一开始处理不需要移动的情况 那么遍历sum[i] 如果往前移动,sum[k]+(sum[i]-sum[i-1])=sum[n]/2,k∈[1,i-1] 如果往后移动,sum[k]-(sum[i]-sum[i-1])=sum[n]/2,k∈[i+1,n] 一开始我没有考虑往后移动 时间复杂度\(O(nlog(n))\) trick 代码 #incl…

这题1<<M为255,可以logN二分答案后,N*M扫一遍表把N行数据转化为一个小于等于255的数字,再255^2检验答案(比扫一遍表复杂度低),复杂度约为N*M*logN #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ][]; ][]; int n,m; int ans,ans2; ],pre[]; int zhuangya(int x,int y){ ;i<=m;++i)…

这题二分下界是0,所以二分写法和以往略有不同,注意考虑所有区间,并且不要死循环... #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; int m,n,k,t; ]; ],y[],z[]; ]; int check(int v){ memset(sum,,sizeof(sum)); ;i<=k;++i) if(z[i]>v) ++sum[x[i]],--sum[y[i]+]; ; ;i…