A - Infinite Sequence
CodeForces - 622A
题意：第一个数是1，接下来是1和2，接下来是1,2, 3，接下来是1,2,3, 4，问第n个数是什么
题解：找出第几轮在找出第几个
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;
const int maxn = 1e2 + 10;
int main()
{
ll n;
while(~scanf("%lld",&n)) {
ll tmp;
tmp = sqrt(2 * n * 1.0);
if(tmp * (tmp + 1) > 2 * n)
tmp--;
if(n - (tmp * (tmp + 1)) / 2 == 0)
printf("%lld\n",tmp);
else
printf("%lld\n",n - (tmp * (tmp + 1)) / 2);
}
}
B - Not Equal on a Segment
CodeForces - 622C
题意：给你一个数组，给你m个询问，每个询问一个区间， 一个数值，在区间里哪一个位置不是这个数值，输出任意一个就行
题解：找到区间里的不一样的两个数字比较一下就好了
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 10;
int a[maxn];
struct node{
int st,en;
int ans;
}b[maxn];
int flag[maxn];
int main()
{
int n, q;
scanf("%d %d", &n, &q);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
int pos = 0;
for (int i = 1; i <= n; i++)
{
if(i == 1)
{
b[pos].st = i;
b[pos].ans = a[i];
}
else
{
if(a[i] != a[i-1])
{
b[pos].en = i - 1;
pos++;
b[pos].st = i;
b[pos].ans = a[i];
}
}
}
b[pos].en = n;
int id = 1;
for(int i=0;i<=pos;i++)
{
for(int j=b[i].st;j<=b[i].en;j++)
flag[j] = id;
id++;
}
// for(int i=0;i<=pos;i++)
// printf("%d %d %d\n",b[i].st,b[i].en,b[i].ans);
// cout<<endl;
// for(int i=1;i<=n;i++)
// printf("%d ",flag[i]);
while(q--)
{
int l,r,x;
scanf("%d %d %d",&l,&r,&x);
if(flag[l] == flag[r] && a[l] == x)
{
puts("-1");
continue;
}
else if(flag[l] == flag[r] && a[r] != x)
{
printf("%d\n",l);
continue;
}
else
{
int id1 = flag[l];
id1--;
if(b[id1].ans != x)
printf("%d\n",l);
else
{
printf("%d\n",b[id1 + 1].st);
}
}
}
}
C - Optimal Number Permutation
CodeForces - 622D
题意：用1-n组长度为2n的序列，要求每个数出现2次，假设位置分别为xi、yi(xi < yi)，定义di = yi-xi。让你找到一种排列使得sigma((n-i) * |di+i-n|)最小。
题解：贪心+找规律
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;
const int maxn = 2e6 + 10;
int a[maxn];
int main()
{
int n;
memset(a,0,sizeof a);
scanf("%d",&n);
int posa = 1;
int posb = n + 1;
for(int i=1;i<n;i++)
{
if(i % 2 == 1)
{
a[posa] = i;
a[n + 1 - posa] = i;
posa++;
}
else
{
a[posb] = i;
a[3 * n - posb] = i;
posb++;
}
}
for(int i = 1;i <= 2 * n;i++)
{
if(a[i] == 0)
printf("%d ",n);
else
printf("%d ",a[i]);
}
}
D - Ants in Leaves
CodeForces - 622E
题意：给定一颗树，每个叶子节点上有一蚂蚁，除了根结点的之外的所有节点任意时刻至多只能有一个蚂蚁，每个蚂蚁每秒能移动到相邻的节点上，问所有蚂蚁移动到根结点的最短时间是多少。
题解：深度小的叶子蚂蚁先走，这样深度小的叶子蚂蚁就不用等深度大的叶子蚂蚁。然后算出每个子树走完的所需时间。d[i] = max(d[i - 1] + 1,d[i])
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;
const int maxn = 5e5 + 10;
int n;
vector<int>g[maxn],d;
void dfs(int u,int fa,int deep)
{
if(g[u].size() == 1)
{
d.push_back(deep);
// printf("%d\n",deep);
}
for(int i=0;i<g[u].size();i++)
{
int v = g[u][i];
if(v == fa)
continue;
dfs(v,u,deep+1);
}
}
int main()
{
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
g[i].clear();
for(int i=1;i<n;i++)
{
int u,v;
scanf("%d %d",&u,&v);
g[u].push_back(v);
g[v].push_back(u);
}
int ans = 0;
for(int k = 0; k < g[1].size(); k++)
{
d.clear();
dfs(g[1][k],1,1);
sort(d.begin(),d.end());
for(int i = 1; i < d.size(); i++)
{
d[i] = max(d[i - 1] + 1,d[i]);
//cout<<d[i]<<endl;
}
ans = max(ans,d.back());
}
printf("%d\n",ans);
}
return 0;
}
E - The Sum of the k-th Powers
CodeForces - 622F
题意：求
mod (1e9 + 7)
原文地址：https://www.cnblogs.com/smallhester/p/11173966.html
时间： 2024-10-11 07:21:51