分成2种情况考虑-男多女少和女多男少
考虑以下2种情况：
BGBGBB
BGBGGG-->GBGBGG 显然后面间隔较多
Program Boys;
var
n,m,i:longint;
begin
assign(input,'input.txt');
assign(output,'output.txt');
reset(input);
rewrite(output);
read(n,m);
if n<m then
begin
for i:=1 to n do write('GB');
for i:=n+1 to m do write('G');
end
else
begin
for i:=1 to m do write('BG');
for i:=m+1 to n do write('B');
end;
writeln;
close(input);close(output);
end.
这题送分题居然wa了
我仔仔细细地看了看题目……发现不慎把Yes 和No打萎了……
仅以为戒
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<cctype>
#include<iostream>
#include<functional>
#include<algorithm>
#include<string>
using namespace std;
#define MAXN (10000+10)
int w[MAXN];
int a[MAXN];
string s;
long long n,t;
int main()
{
// freopen("read.in","r",stdin);
// freopen("read.out","w",stdout);
cin>>t>>n;
for (int i=(int)('a');i<=(int)('z');i++) {
cin>>w[i];
}
// cout<<w[112];
for (int i=1;i<=n;i++)
{
cin>>s;
a[i]=0;
for (int j=0;j<s.length();j++)
{
s[j]=tolower(s[j]);
// cout<<w[(tolower(s[i]))]<<' ';
a[i] =a[i]+ w[s[j]];
// cout<<a[i]<<' '<<endl;
}
}
sort(a+1,a+1+n);
// for (int i=1;i<=n;i++) cout<<a[i]<<' ';
for (int i=1;i<=n;i++)
{
t-=(long long)(a[i]);
if (t<(long long)0)
{
cout<<"Non";
cout<<(i-1)<<endl;
return 0;
}
}
cout<<"Yesn"<<t<<endl;
return 0;
}
题目大意：是一堆人来h点m分来超市买东西，同时可以有an位顾客买单，买单可认为1分钟以内完成，问至少有几位售货员才能使所有顾客不等待
直接统计……
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<functional>
using namespace std;
#define MAXN (100000+10)
int n,i;
int a[MAXN]={0};
int main()
{
int ans=0;
scanf("%d",&n);
for (int i=1;i<=n;i++)
{
int h,m;
scanf("%d%d",&h,&m);
a[h*60+m]++;
ans=max(ans,a[h*60+m]);
}
printf("%dn",ans);
// while (1);
return 0;
}
这题枚水过……
var
s1,s2,s3:longint;
i,j,k:longint;
f:array[0..100] of longint;
begin
read(s1,s2,s3);
fillchar(f,sizeof(f),0);
for i:=1 to s1 do
for j:=1 to s2 do
for k:=1 to s3 do
begin
inc(f[i+j+k]);
end;
j:=1;
for i:=1 to 100 do
if f[i]>f[j] then j:=i;
writeln(j);
end.
暴力模拟无算法
Program P1868;
Var
c:char;
n,i,j,k:longint;
a:array[0..10000] of longint;
function is_ant:boolean;
var
i,j,k:longint;
begin
for k:=1 to n shr 2 do
for i:=1 to n-k shl 1-1 do
if ((a[i]<a[i+k]) and (a[i+k]<a[i+2*k])) or ((a[i]>a[i+k]) and (a[i+k]>a[i+2*k])) then exit(false);
exit(true);
end;
Begin
while true do
begin
read(c);
if c='0' then break;
val(c,n);
repeat
read(c);
if c=':' then break;
n:=n*10+ord(c)-48;
until false;
for i:=1 to n do
begin
read(j);
a[j]:=i;
end;
if is_ant then writeln('yes') else writeln('no');
readln;
end;
end.
求余数……
Program P3980;
var
a,b:longint;
begin
while not seekeof do
begin
read(a,b);
writeln(a mod b);
end;
end.
这题的数据太水……
居然不需要2分Log n查找……尽管我也不知道上限是多少……
Program P3100;
var
a,b,n,i,j,dis:longint;
function pow(a,b:longint):longint;
var
i:longint;
begin
if a=0 then exit(0);
if a=1 then exit(1);
if b=0 then exit(0);
if b=1 then exit(a)
else
begin
pow:=pow(a,b shr 1);
if (b mod 2=0) then
begin
exit(pow*pow);
end
else exit(pow*pow*a);
end;
end;
begin
read(b,n);
while (b+n>0) do
begin
i:=0;
while (pow(i,n)<b) do
begin
dis:=b-pow(i,n);
inc(i);
end;
if (pow(i,n)-b>dis) then writeln(i-1) else writeln(i);
read(b,n);
end;
end.
这题题目看不懂……用谷歌翻译完，更看不懂……
对A此题不看Discuss的神语言流表示膜拜
Program P2209;
var
n,p,s,ans,i:longint;
begin
read(n,p);
ans:=0;
for i:=1 to n do
begin
read(s);
if p=2 then s:=s*s;
if p=3 then s:=s*s*s;
if s>0 then inc(ans,s);
end;
writeln(ans);
end.
别问我……我也不知道怎么证明
Program p2505;
var
n:double;
begin
while not seekeof do
begin
read(n);
while n>18 do n:=n/18;
if n<=9 then writeln('Stan wins.')
else writeln('Ollie wins.');
end;
end.
3n+1问题 果断暴力
Program P1207;
var
i,j,k,n,m,ans:longint;
function max(a,b:longint):longint;
begin
if a>b then exit(a) else exit(b);
end;
procedure swap(var a,b:longint);
var
p:longint;
begin
p:=a;
a:=b;
b:=p;
end;
begin
while not eof do
begin
readln(n,m);
write(n,' ',m,' ');
if n>m then swap(n,m);
ans:=0;
for i:=n to m do
begin
j:=1;
k:=i;
while (k<>1) do
begin
if (k mod 2=0) then k:=k div 2
else k:=k*3+1;
inc(j);
end;
ans:=max(ans,j);
end;
writeln(ans);
end;
end.