codechef是真敢给分，上把刚注册，这把就div2了，再加上一周没打过还是有点不适应的，好在最后还是能够顺利上分
今天的封面是P3R的设置菜单 我抠出来做我自己的游戏主页了（
A - Convert string
题意
在01串里面可以翻转k次，问从1到n有多少个k有机会使01串所有元素相同
思路
暴力遍历求解即可，一开始下标打错了wa了一发
代码
cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
void solve()
{
int n;
scanf("%d",&n);
string s;
cin>>s;
int one=0,zero=0;
for(int i=0;i<n;i++)
{
if(s[i]=='1')
{
one++;
}
else
{
zero++;
}
}
int cnt=0;
for(int k=1;k<=n;k++)
{
//全变1
if(k>=zero and (k-zero)%2==0)
{
// printf("%d ",k);
cnt++;
continue;
}
//全变0
if(k>=one and (k-one)%2==0)
{
// printf("%d ",k);
cnt++;
continue;
}
}
// printf("\n");
printf("%d\n",cnt);
}
int main()
{
int T=1;
scanf("%d",&T);
while(T--)
{
solve();
}
return 0;
}B - Ball Game
题意
好多小球冲家，速度各不相同，中间撞上了就会融合成一个速度取最大值的新小球，问最后一共有多少个小球到终点
思路
所谓的融合其实就是大球吃小球，维护个pq然后大小比较无效化就行
代码
cpp
#include <iostream>
#include <algorithm>
#include <map>
#include <cstring>
#include <string>
#include <vector>
#include <cctype>
#include <cmath>
#include <queue>
#include <set>
using namespace std;
typedef long long LL;
typedef pair<LL, int> PLL;
#define qwq ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define tcase \
LL t = read(); \
while (t--)
LL read()
{
LL sum = 0, fl = 1;
int ch = getchar();
for (; !isdigit(ch); ch = getchar())
if (ch == '-')
fl = -1;
for (; isdigit(ch); ch = getchar())
sum = sum * 10 + ch - '0';
return sum * fl;
}
const int N = 1e6 + 10;
struct node
{
LL d, v;
bool operator<(const node &T) const
{
return d < T.d;
}
} a[N];
struct node2
{
LL x, y;
bool operator<(const node2 &T) const
{
return x * T.y < T.x * y;
}
};
void solve()
{
LL n = read();
for (int i = 1; i <= n;i++)
a[i].d = read();
for (int i = 1; i <= n;i++)
a[i].v = read();
sort(a + 1, a + 1 + n);
priority_queue<node2> pq;
for (int i = 1; i <= n;i++)
{
node2 nowt = {a[i].d,a[i].v};
while(pq.size()&&nowt<pq.top())
pq.pop();
pq.push(nowt);
}
printf("%lld\n", pq.size());
}
int main()
{
tcase
solve();
return 0;
}C - Sequence Search
题意
X 1 = 0 ; X_1 = 0; X1=0;
X 2 = A ; X_2 = A; X2=A;
X 3 = B ; X_3 = B; X3=B;
X i = X i − 1 + X i − 2 = X i − 3 ; X_i = X_{i-1}+X_{i-2}=X_{i-3}; Xi=Xi−1+Xi−2=Xi−3;
一直A和B，求数组中第k小的数
思路
简单枚举一下很容易可以发现规律
bash
0,A,B,A+B,2B,A+2B,3B,A+3B,4B,A+4B,5B,A+5B,6B,A+6B,7B,A+7B,8B,A+8B,9B,A+9B,10B,A+10B,11B,A+11B,12B,A+12B,13B,A+13B,14B然后现在还要排序，很容易想到比较时一个很重要的印象因素就是A是B的几倍，然后再枚举下找下规律就行了
bash
0,B,A,2B,A+B,3B,A+2B,4B,A+3B,5B A/B==1
0,B,2B,A,3B,A+B,4B A/B==2代码
cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
//0,A,B,A+B,2B,A+2B,3B,A+3B,4B,A+4B,5B,A+5B,6B,A+6B,7B,A+7B,8B,A+8B,9B,A+9B,10B,A+10B,11B,A+11B,12B,A+12B,13B,A+13B,14B,A+14B,15B,A+15B,16B,A+16B,17B,A+17B,18B,A+18B,19B,A+19B,20B,A+20B,21B,A+21B,22B,A+22B,23B,A+23B,24B,A+24B,25B,A+25B,26B,A+26B,27B,A+27B,28B,A+28B,29B,A+29B,30B,A+30B,31B,A+31B,32B,A+32B,33B,A+33B,34B,A+34B,35B,A+35B,36B,A+36B,37B,A+37B,38B,A+38B,39B,A+39B,40B,A+40B,41B,A+41B,42B,A+42B,43B,A+43B,44B,A+44B,45B,A+45B,46B,A+46B,47B,A+47B,48B,A+48B,49B,A+49B,50B,A+50B,51B,A+51B,52B,A+52B,53B,A+53B,54B,A+54B,55B,A+55B,56B,A+56B,57B,A+57B,58B,A+58B,59B,A+59B,60B,A+60B,61B,A+61B,62B,A+62B,63B,A+63B,64B,A+64B,65B,A+65B,66B,A+66B,67B,A+67B,68B,A+68B,69B,A+69B,70B,A+70B,71B,A+71B,72B,A+72B,73B,A+73B,74B,A+74
//0,B,A,2B,A+B,3B,A+2B,4B,A+3B,5B A/B==1
//0,B,2B,A,3B,A+B,4B A/B==2
void solve()
{
ll A,B,K;
scanf("%lld%lld%lld",&A,&B,&K);
int time=A/B;
int start=3+time;
if(K==start-1)
{
printf("%lld\n",A);
return ;
}
if(K==1)
{
printf("%lld\n",0);
return ;
}
if(K<start)
{
printf("%lld\n",(K-1)*B);
return ;
}
K-=start;
K++;
if(K%2==1)
{
printf("%lld\n",((K+1)/2+time)*B);
return ;
}
else
{
printf("%lld\n",(K/2)*B+A);
return ;
}
}
int main()
{
int T=1;
scanf("%d",&T);
while(T--)
{
solve();
}
return 0;
}D - Shooting (Easy)
题意
射飞镖，以一个点为中心点，距离为得分，求两个人在每个点作为中心点情况得分的差值
思路
前缀和维护即可
代码
cpp
#include <iostream>
#include <algorithm>
#include <map>
#include <cstring>
#include <string>
#include <vector>
#include <cctype>
#include <cmath>
#include <queue>
#include <set>
using namespace std;
typedef long long LL;
typedef pair<LL, int> PLL;
#define qwq ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define tcase \
LL t = read(); \
while (t--)
LL read()
{
LL sum = 0, fl = 1;
int ch = getchar();
for (; !isdigit(ch); ch = getchar())
if (ch == '-')
fl = -1;
for (; isdigit(ch); ch = getchar())
sum = sum * 10 + ch - '0';
return sum * fl;
}
const int N = 1e6 + 10;
LL d[N];
LL s1[N], s2[N];
void solve()
{
LL sum1 = 0, sum2 = 0;
LL n = read(),m = read();
n = m;
for (int i = 1; i <= m;i++)
d[i] = read();
for (int i = 1; i <= m;i++)
{
s1[i] = s1[i - 1] + 1;
s2[i] = s2[i - 1] + 1;
if(d[i]==1)
{
s1[i]++;
sum1 += i - 1;
}
else if(d[i]==2)
{
s2[i]++;
sum2 += i - 1;
}
else if(d[i]==3)
{
s1[i]++;
s2[i]++;
sum1 += i - 1;
sum2 += i - 1;
}
// printf("%lld %lld\n", s1[i], s2[i]);
}
//printf("%lld %lld\n", s1[m], s2[m]);
printf("%lld ", llabs(sum1 - sum2));
// printf("de %lld %lld\n", sum1, sum2);
for (int i = 2; i <= m;i++)
{
sum1 += s1[i - 1];
sum1 -= s1[m] - s1[i - 1];
// printf("add:%lld sub:%lld\n", s1[i - 1], s1[m] - s1[i - 1]);
sum2 += s2[i - 1];
sum2 -= s2[m] - s2[i - 1];
// printf("add:%lld sub:%lld\n", s2[i - 1], s2[m] - s2[i - 1]);
printf("%lld ", llabs(sum1 - sum2));
// printf("de %lld %lld\n", sum1 , sum2);
}
printf("\n");
}
int main()
{
tcase
solve();
return 0;
}