116 条题解
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芒果木瓜榴莲 LV 9 @ 2009-11-04 11:10:59
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├ 测试数据 01：答案正确... 0ms
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Accepted 有效得分：100 有效耗时：0msvar n,p,x,y,z:longint;
begin
readln(n);
for p:=1 to n do
begin
readln(x,y,z);
if ((y=9)and(z=30))or((y=11)and(z=30))or(not(odd(y+z)))then writeln('YES') else writeln('NO');
end;
end.奇偶性的题解很好，希望考试的时候能想到
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编译通过...
├ 测试数据 01：答案正确... 0ms
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Accepted 有效得分：100 有效耗时：0msprogram ex;
const days:array[1..12]of longint=(31,28,31,30,31,30,31,31,30,31,30,31);
var i,j,n:longint;
year:array[1900..2010]of boolean;
f:array[1900..2010,1..12,1..31]of boolean;
rest:array[1..12]of longint;
bo:boolean;procedure check;
var i,j:longint;
begin
fillchar(year,sizeof(year),false);
for i:=1900 to 2010 do
if ((i mod 4=0)and(i mod 1000))or(i mod 400=0) then
year[i]:=true;
end;procedure main;
var i,j,y,m,d,y1,y2,m1,m2,d1,d2,m0,mt:longint;
begin
fillchar(rest,sizeof(rest),0);
fillchar(f,sizeof(f),false);
f[2001,11,3]:=true;
f[2001,11,1]:=true;
for i:=4 to 31 do
if (not odd(i)) then
f[2001,10,i]:=true;
f[2001,10,2]:=true;
for y:=2001 downto 1900 do
begin
if year[y] then rest[2]:=1 else rest[2]:=0;
if y=2001 then m0:=9 else m0:=12;
for m:=m0 downto 1 do
for d:=days[m]+rest[m] downto 1 do
begin
y1:=y;m1:=m;d1:=d+1;
if d1>days[m]+rest[m] then
begin
d1:=1;
inc(m1);
if m1>12 then
begin
m1:=1;
inc(y1);
end;
end;
if m+1>12 then mt:=1 else mt:=m+1;
if days[mt]+rest[mt]>=d then
begin
y2:=y;m2:=m+1;d2:=d;
if m2>12 then
begin
m2:=1;
inc(y2);
end;
end
else
begin
y2:=y1;m2:=m1;d2:=d1;
end;
f[y,m,d]:=(not((f[y1,m1,d1]) and (f[y2,m2,d2])));
end;
end;
end;procedure indoit;
var i,j,x,y,z:longint;
begin
readln(n);
for i:=1 to n do
begin
readln(x,y,z);
if f[x,y,z] then writeln('YES') else writeln('NO');
end;
end;begin
check;
main;
indoit;
end.写完这个。。。 我只想吐血。。。。
看不懂数学方法 所以写了模拟
模拟+状态转移 -
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ly19920507 LV 8 @ 2009-10-08 11:39:59
这题数据有问题啊。。。。。
我交程序的时候想到了错误，怎么就A了？
点开一看，哇，只有1个测试点 -
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...此题...很让人恼火...居然是变到2006.11.4才算赢...
2001.11.4的话,WA...郁闷呐...
感觉我这个代码比较好理解...
就贴了一下...
f[y,m,d]表示此日期是否必胜.
状态转移:f[y,m,d]:=not(f[y,m+1,d] and f[y,m,d+1]);
program voj1004;
const md:array[1..12]of longint=(31,28,31,30,31,30,31,31,30,31,30,31);
var f:array[1900..2010,1..12,1..31]of boolean;
year:array[1900..2010]of boolean;
r:array[1..12]of longint;procedure start;
var i:longint;
begin
fillchar(year,sizeof(year),false);
for i:=1900 to 2010 do
if (i mod 4=0) and not ((i mod 100=0) and (i mod 4000)) then
year[i]:=true;
fillchar(f,sizeof(f),true);
f[2006,11,4]:=false;
f[2006,11,2]:=false;
end;procedure main;
var y,m,d,y1,m1,d1,y2,m2,d2,m0:longint;
begin
for y:=2006 downto 1900 do begin
if year[y] then r[2]:=1 else r[2]:=0;
if y=2006 then m0:=10 else m0:=12;
for m:=m0 downto 1 do
for d:=md[m]+r[m] downto 1 do begin
y1:=y; m1:=m; d1:=d+1;
if d1>md[m]+r[m] then begin
d1:=1; inc(m1);
if m1>12 then begin
m1:=1; inc(y1);
end;
end;
y2:=y; m2:=m+1; d2:=d;
if m2>12 then begin
m2:=1; inc(y2);
end;
f[y,m,d]:=not (f[y1,m1,d1] and f[y2,m2,d2]);
end;
end;
end;procedure writeout;
var i,a,b,c:longint;
begin
readln(i);
for i:=1 to i do begin
readln(a,b,c);
if f[a,b,c] then writeln('YES') else writeln('NO');
end;
end;begin
start;
main;
writeout;
end. -
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关于奇偶我来解释……
我们设2001.11.4必败，推上去，即2001.10.4和2001.11.3必胜（此时m+d=偶数）再推上去是11.2，10.3，9.3，都是必败（此时m+d=奇数），以此类推……
为什么9.30和11.30例外呢？
因为9.30和11.30月份+1，即10.30和12.30，都是必败局面
这种情况只有在8月与以后出现，为什么大家应该清楚
2.29为奇数，必败
没有2.29，剩2.28和3.1衔接也没问题，都是必胜
12.31必败，1.1必胜（以上情况自己考虑）
所以本题和年份完全没关系-
zone @ 2016-07-27 10:04:42
Orz
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const
m:array[1..13] of integer=(31,28,31,30,31,30,31,31,30,31,30,31,0);
var
a:array[1..30011231] of byte;
i,j,k,n:longint;
s:string;
year,month,day,g,p,q,x,y,z,get1,get2:longint;
begin
year:=2001; month:=11; day:=4;
for i:=20011105 to 20011231 do
a[i]:=1;
while (year1900) or (month1) or (day1) do
begin
dec(day);
if day=0 then
begin
dec(month);
if month = 0 then
begin
dec(year);
month:=12;
end;
day:=m[month];
if (month=2) and (year mod 4 = 0) then inc(day);
end;
g:=year*10000+month*100+day;
{get1}
y:=month+1;
x:=m[y]; if (y=2) and (year mod 4 = 0) and (year1900) then x:=x+1;
if (y=day) then
get1:=a[year*10000+(month+1)*100+day]
else if (y=13) then
get1:=a[(year+1)*10000+100+day]
else
get1:=1;
{get2}
y:=day+1; x:=month; z:=year;
p:=m[x]; if (year mod 4 = 0) and (x=2) and (year1900) then inc(p);
if p -
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一开始说我存取非法。。。
我的数组从 1900 01 01 到 2001 12 31
居然不够大！！改成 1 ~ 3001 12 31
就过了！~~
AC
const
m:array[1..13] of integer=(31,28,31,30,31,30,31,31,30,31,30,31,0);
var
a:array[1..30011231] of byte;
i,j,k,n:longint;
s:string;
year,month,day,g,p,q,x,y,z,get1,get2:longint;
begin
year:=2001; month:=11; day:=4;
for i:=20011105 to 20011231 do
a[i]:=1;
while (year1900) or (month1) or (day1) do
begin
dec(day);
if day=0 then
begin
dec(month);
if month = 0 then
begin
dec(year);
month:=12;
end;
day:=m[month];
if (month=2) and (year mod 4 = 0) then inc(day);
end;
g:=year*10000+month*100+day;
{get1}
y:=month+1;
x:=m[y]; if (y=2) and (year mod 4 = 0) and (year1900) then x:=x+1;
if (y=day) then
get1:=a[year*10000+(month+1)*100+day]
else if (y=13) then
get1:=a[(year+1)*10000+100+day]
else
get1:=1;
{get2}
y:=day+1; x:=month; z:=year;
p:=m[x]; if (year mod 4 = 0) and (x=2) and (year1900) then inc(p);
if p -
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//我崩溃了。。。。总是说我超时。。。
#include
using namespace std;
void fun(int *year,int *month,int *day);//变换一次日期
int main()
{
int i,n,y[100],m[100],d[100],x[100]={0};
cin>>n;
for(i=0;i>y[i]>>m[i]>>d[i];
while((y[i]==2006&&m[i]==11&&d[i]==4)==0)
{
fun(&y[i],&m[i],&d[i]);
x[i]++;
}
}
for(i=0;i -
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如对本题有疑问可以参看我的题解：http://xujieqi.blog.hexun.com/35722312_d.html
芒果木瓜榴莲
LV 9
@ 2009-11-04 11:10:59
