一开始网站爆炸,502刷新了10min,那时就知道这场要gg了。。。
A.签到题
B.求一个01串包含a个0,b个1且恰好有x个位置s[i]不等于s[i+1];
考虑a,b大小关系确定第一个是0还是1,前面构造x-1个10或01,后面就连着输出相同的就好了。
没看到样例wa3两次。。。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define ll long long
#define out(a) printf("%d",a)
#define writeln printf("\n")
#define min(a,b) a<b?a:b;
#define max(a,b) a>b?a:b;
using namespace std;
int n,m,a,b,c,d,x,ans,now;
int read()
{
int s=0,t=1; char c;
while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();}
while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();}
return s*t;
}
ll readl()
{
ll s=0,t=1; char c;
while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();}
while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();}
return s*t;
}
int main()
{
b=read(); a=read(); x=read();
if (x==1) {
for (int i=1;i<=a;i++)
out(1);
for (int i=1;i<=b;i++)
out(0);
return 0;
}
if (a>b) {
for (int i=1;i<=x;i++)
if (i&1) out(1),c++,now=1;
else out(0),d++,now=0;
}
else {
for (int i=1;i<=x;i++)
if (i&1) out(0),d++,now=0;
else out(1),c++,now=1;
}
a-=c; b-=d;
if (now==1) {
for (int i=1;i<=a;i++)
out(1);
for (int i=1;i<=b;i++)
out(0);
}
else {
for (int i=1;i<=b;i++)
out(0);
for (int i=1;i<=a;i++)
out(1);
}
return 0;
}
C.找一个不小于k的区间平均值最大.
读错题意罚时爆炸QAQ。
记录前缀和然后二重循环扫一遍即可。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define ll long long
#define out(a) printf("%lld",a)
#define writeln printf("\n")
#define min(a,b) a<b?a:b;
#define max(a,b) a>b?a:b;
using namespace std;
int n,m,a,b,x,k,num;
double ans;
int sum[100050];
int read()
{
int s=0,t=1; char c;
while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();}
while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();}
return s*t;
}
ll readl()
{
ll s=0,t=1; char c;
while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();}
while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();}
return s*t;
}
int main()
{
n=read(); k=read();
for (int i=1;i<=n;i++) {
x=read(); sum[i]=sum[i-1]+x;
}
for (int j=k;j<=n;j++) {
num=-2333333;
for (int i=j;i<=n;i++)
num=max(num,sum[i]-sum[i-j]);
if ((double)num/j>ans) ans=(double)num/j;
}
printf("%.15f",ans);
return 0;
}
D.给n个2的d次方的数,m个询问问是否有n个数的子集和等于x.
正解都想出来了,可就是没写出来,细节总是错错错QAQ
考虑把所有数都搞成2进制,如果这一位不够的话就往下一位拆。。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define ll long long
#define out(a) printf("%d",a)
#define writeln printf("\n")
using namespace std;
int n,k;
int x,cnt,m,tot,ans;
int a[200050],b[200050];
bool flag;
int read()
{
int s=0,t=1; char c;
while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();}
while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();}
return s*t;
}
ll readl()
{
ll s=0,t=1; char c;
while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();}
while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();}
return s*t;
}
int main()
{
n=read(),k=read();
for (int i=1;i<=n;i++){
x=read(); cnt=0;
while (true) {
if (x==1||x==0) break;
cnt++; x/=2;
}
a[cnt]++; //out(cnt); writeln;
}
for (int i=1;i<=k;i++){
m=read(); flag=false; tot=ans=0; b[0]=0;
while (true) {
b[++tot]=m%2;
if (flag) break;
m/=2;
if (m==0||m==1) flag=true;
}
for (int j=tot;j>=0;j--)
if (a[j-1]>=b[j]) ans+=b[j],b[j]=0;
else {
b[j]-=a[j-1];
b[j-1]+=b[j]*2;
ans+=a[j-1];
}
if (b[0]>0) out(-1);
else out(ans);
writeln;
}
return 0;
}
总结
1.码力还是太差了,细节问题总出错。
2.心态问题,急忙交题不测样例是大忌,被一些bug卡的很烦。
3.读题能力差,现在由于cf复制的问题不太兹磁翻译,所以基本上都是读英语或dalao翻译,题意总是看错。
There is still a long way to go....
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