ZZU_LGY

题目链接#include <iostream>#include <cstring>#include <algorithm>#include <vector>using namespace std;#define int long longvoid solve(){ int n,c; cin>>n>>c; vector<int>val(c+1); for(int i=1,cos

题目链接#include <iostream>#include <cstring>#include <algorithm>#include <vector>using namespace std;#define int long longconst int N = 400010;int s[N];int tr[N];int n,m;vector<int>nums;int lowbit(int x){ return.

题目链接自己二分写挂了，得了90分，不知道哪里错了，参考大佬简洁代码#include <iostream>#include <cstring>#include <algorithm>#include <vector>#include <functional>#include <set>using namespace std;typedef pair<int,int> PII;#define int lo

这题傻了，用dp写了俩多小时没写出来。。。#include <iostream>#include <cstring>#include <algorithm>using namespace std;#define int long longconst int N = 1000010,INF=0x3f3f3f3f;char s[N];int n,k,w,q;signed main(){ scanf("%lld",&n); sca..

这题双指针没想出来，用二分+前缀和写出来了，写的巨复杂，下面是两种写法二分+前缀和写法#include <iostream>#include <cstring>#include <algorithm>#include <vector>#include <map>using namespace std;#define int long longconst int N = 200010;char s[N];int n,k;in.

D.Palindromes Coloring思路：题意是在一个字符串中选出k个子序列，将子序列任意排序，使每个子序列是回文的，求子序列中长度最小值的最大值。一道贪心题，要想回文，至多有一个字母出现奇数次，其余字母均出现偶数次，因此答案只与字母出现次数有关，跟字母是谁没关系。统计出现偶数次字母的总次数，即sum+=x/2答案 ans满足ans/2*k<=sum可以二分来写。。但是小写字母只有26个，因此字符串很长的话，答案一定很大，所以遍历也能过。。。#include <iostr

题目链接#include <iostream>#include <cstring>#include <algorithm>#include <queue>#include <map>using namespace std;const int N = 100010;#define int long longtypedef pair<int,int>PII;int v,n,m;PII a[N];int s1[N]

题目链接#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int N = 510;//#define int long longint n,m,A,B;int s[N][N],a[N][N];bool check(int x){ int cnta=0,cntb=0,fa=1,fb=1,sum=0; for(int i

题目链接#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int N = 5000010;#define int long longint n,sa,sb,sc,sd,a1,mod;int a[N]={0},b[N];int f(int x){ return sa*x%mod*x%mod*x%mod+sb*x%mod*x%mo

题目链接#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int N = 100010;#define int long longint w[10]={10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};int n,a[N];int cal(int l,int

题目链接#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int N = 100010;#define int long longint a[N],b[N],c[N];int n,k;int check(int mid){ for(int i=1;i<=n;i++)c[i]=b[i]-a[i]*mid; sort(c

题目链接这道二分题特别之处在于评判标准比较特殊，是求范围最接近的值#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int N = 200010;#define int long longint n,m,s;int L[N],R[N],w[N],v[N];int check(int mid){ int cnt1[N]={0},c

K-th Number题意：从长度为n的区间中，找出长度大于等于k的区间的第k大值，把这些值放在B数组中，求B中第m大值思路：容易往树状数组和线段树那里思考，但其实是二分设 mid ⩽\leqslant⩽ ans有在B中，mid为第>=m的值，即在B中比mid大的值数量>=m,即在A的长度大于k的子区间中，比mid大的第k大的值的个数>=m个，即有>=m个子区间，每个区间中，比mid大的值的个数大于k个所以有for(int i=1;i<=n;i++)s[i]

题目链接#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int N = 1000010;#define int long longint n,m;int d[N],s[N],t[N],a[N],b[N];bool check(int mid){ for(int i=1;i<=n;i++)b[i]=a[i]-a[i-1]

题目链接#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int N = 55;#define int long longint n,m;int a[N];bool check(int mid){ if(mid>a[2]+a[1])return false; int sum=0; for(int i=1;

题目链接#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int N = 100010;#define int long longint a[N],id[N];int n,d;bool check(int mid){ int sum=0,cnt=0; for(int i=1;i<=d;i++) {

题目链接#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int N = 100010;int a[N];int n;bool check(int mid){ int s[N]={0},sum=0; for(int i=1;i<mid;i++)s[a[i]]++; for(int i=1;i<mid;