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Turbo Pascal提供了八个标准函数和标准过程，利用它们可以灵活解决字符串解题中的一些问题。
length(st:string):byte 函数，返回st串的长度，既串中字符的个数。函数值字节型byte 0～255。 pos(sub,st:string):byte 函数，在st串中找子串sub，找到后返回值为sub在st中的位置，若没找到,函数值为0。 str(value,st) 函数，将整数或实数value转换成数字字符串st 。 val(st, value,code) 过程，将字符串st 转换成数value, code返回st中第一个非法字符的位置，未有错，code为0。 copy(st,pos,num) 函数, 在字符串st中从第pos个字符开始顺序截取num个字符。若pos大于st的长度，则返回空串。 insert(sub,st,pos) 过程，在字符串st的第pos个字符位置处插入子串sub。 delete(st,pos,num) 过程，删除st中第pos个字符开始的num个字符。 upcase(st):st 函数，将字母ch转换成大写字母。 |
Author Archives: nike0good
POJ 1008(模拟）
最后一天特判
Program P1008;
var
n,i,j1,j2,j,day,month,year:longint;
daytotal,tday,tmonth,tyear:longint;
s,s1:string;
mon:array[1..19] of string=('pop','no','zip','zotz','tzec','xul','yoxkin','mol','chen','yax','zac','ceh','mac','kankin','muan', 'pax', 'koyab', 'cumhu','uayet');
tmon:array[1..20] of string=('imix', 'ik', 'akbal','kan','chicchan','cimi','manik','lamat','muluk','ok','chuen','eb','ben','ix','mem','cib','caban','eznab','canac','ahau');
begin
readln(n);
writeln(n);
for i:=1 to n do
begin
readln(s);
j1:=1;
while (s[j1]<>'.') do inc(j1);
s1:=copy(s,1,j1-1);
val(s1,day);
j2:=length(s);
while (s[j2]<>' ') do dec(j2);
s1:=copy(s,j2+1,5);
val(s1,year);
inc(j1);
s1:=copy(s,j1+1,j2-j1-1);
for j:=1 to 19 do
if s1=mon[j] then
begin
month:=j;
break;
end;
daytotal:=year*365+day+1+(month-1)*20;
tyear:=(daytotal-1) div 260;
tday:=(daytotal mod 13);
if tday=0 then tday:=13;
tmonth:=(daytotal mod 20);
if tmonth=0 then tmonth:=20;
writeln(tday,' ',tmon[tmonth],' ',tyear);
end;
end.
POJ 1836（双向LIS)
双向LIS……居然数组又开小了……
Program P1836;
var
n,i,j,ans:longint;
a:array[1..1000] of double;
ll,lr:array[1..1000] of longint;
function min(a,b:longint):longint;
begin
if a<b then exit(a) else exit(b);
end;
begin
fillchar(ll,sizeof(ll),0);
fillchar(lr,sizeof(lr),0);
read(n);
for i:=1 to n do read(a[i]);
ll[1]:=1;
for i:=2 to n do
begin
for j:=1 to i-1 do
if a[j]<a[i] then if (ll[j]>=ll[i]) then
ll[i]:=ll[j]+1;
if ll[i]=0 then inc(ll[i]);
end;
lr[n]:=1;
for i:=n-1 downto 1 do
begin
for j:=i+1 to n do
if a[j]<a[i] then if (lr[j]>=lr[i]) then
lr[i]:=lr[j]+1;
if lr[i]=0 then inc(lr[i]);
end;
ans:=n;
for i:=1 to n do
ans:=min(ans,n-(ll[i]+lr[i]-1));
for i:=1 to n do
for j:=i+1 to n do
ans:=min(ans,n-(ll[i]+lr[j]));
writeln(ans);
end.
POJ 1837(Dp水题）
这题居然不用高精度就能过……测试数据好弱
Program P1837;
var
c,g,i,j,k,p:longint;
li,w:array[1..20] of longint;
f:array[1..20,-7500..7500] of longint;
begin
fillchar(f,sizeof(f),0);
read(c,g);
for i:=1 to c do read(li[i]);
for i:=1 to g do read(w[i]);
for i:=1 to c do f[1,li[i]*w[1]]:=1;
for i:=2 to g do
for j:=1 to c do
begin
p:=li[j]*w[i];
if p<0 then
begin
for k:=-7500 to 7500+p do
if f[i-1,k-p]>0 then
inc(f[i,k],f[i-1,k-p])
end
else
begin
for k:=-7500+p to 7500 do
if f[i-1,k-p]>0 then
inc(f[i,k],f[i-1,k-p]);
end;
end;
writeln(f[g,0]);
end.
POJ 1007(求逆序对数）
求逆序对数，2关键字排序
Program P1007;
var
n,m,i,j,k,l,p:longint;
a:array[1..200] of string;
s:string;
b,num:array[1..200] of longint;
function h(s:string):longint;
var
i,j,a,c,g,t:longint;
begin
a:=0;
c:=0;
g:=0;
t:=0;
h:=0;
for i:=length(s) downto 1 do
begin
if s[i]='A' then inc(a);
if s[i]='C' then
begin
inc(c);
inc(h,a);
end;
if s[i]='G' then
begin
inc(g);
inc(h,a+c);
end;
if s[i]='T' then
begin
inc(t);
inc(h,a+c+g);
end;
end;
end;
procedure qsort(l,r:longint);
var
i,j,m,p:longint;
s2:string;
begin
i:=l;
j:=r;
m:=b[(l+r) div 2];
repeat
while b[i]<m do inc(i);
while b[j]>m do dec(j);
if i<=j then
begin
p:=b[i];
b[i]:=b[j];
b[j]:=p;
p:=num[i];
num[i]:=num[j];
num[j]:=p;
{ s2:=a[i];
a[i]:=a[j];
a[j]:=s2;
} inc(i);
dec(j);
end;
until i>j;
if l<j then qsort(l,j);
if i<r then qsort(i,r);
end;
begin
readln(n,m);
for i:=1 to m do
begin
readln(a[i]);
b[i]:=h(a[i]);
num[i]:=i;
end;
qsort(1,m);
i:=1;
while (i<m) do
begin
if b[i]<b[i+1] then inc(i)
else
begin
for j:=i+1 to m do if b[j]<>b[i] then break;
if (b[i]<>b[m]) then dec(J);
for k:=i to j-1 do
for l:=k+1 to j do
if num[k]>num[l] then
begin
p:=num[k];
num[k]:=num[l];
num[l]:=p;
{ s:=a[k];
a[k]:=a[l];
a[l]:=s;
} end;
i:=j+1;
end;
end;
for i:=1 to m do writeln(a[num[i]]);
end.
POJ 1006(中国剩余定理）
中国剩余定理
若一个数除m1余p1,除m2余p2……,除mn余pn （m1,m2……,mn互质）
则求 k1使k1=m2*……*mn的倍数且除m1余1
……
则这个数为(k1*p1+k2*p2+……kn*pn) mod (m1*m2*……*mn)
Program P1005;
var
a,b,c,d,i,a1,a2,a3,ans:longint;
begin
i:=1;
a1:=28*33;
a2:=23*33;
a3:=23*28;
while (a1 mod 23<>1) do inc(a1,28*33);
while (a2 mod 28<>1) do inc(a2,23*33);
while (a3 mod 33<>1) do inc(a3,23*28);
while (true) do
begin
read(a,b,c,d);
if (a=-1) and (b=-1) and (c=-1) and (d=-1) then break;
ans:=(a1*a+a2*b+a3*c+21252-d) mod 21252;
if ans=0 then ans:=21252;
writeln('Case ',i,': the next triple peak occurs in ',ans,' days.');
inc(i);
end;
end.
POJ 1005(向后去整）
小数处理……
Program P1005;
const
pi=3.1415926;
Var
t,i,j:longint;
x,y,s:double;
begin
read(t);
for i:=1 to t do
begin
read(x,y);
s:=sqrt(x*x+y*y);
s:=s*s*pi/2;
s:=s/50;
if trunc(s)<s then s:=trunc(s)+1
else s:=trunc(s);
writeln('Property ',i,': This property will begin eroding in year ',s:0:0,'.');
end;
writeln('END OF OUTPUT.');
end.
POJ 1004(小数处理）
输出巨恶心，行末无回车
Program P1004;
var
a,b:double;
i:longint;
begin
read(a);
for i:=2 to 12 do
begin
read(b);
a:=a+b;
end;
a:=a/12;
write('$',a:2:2);
end.
POJ 1014(背包问题）
限制背包
Program P1014;
const
maxv=120000;
n=6;
var
a:array[1..6] of longint;
i,j,k,l,v:longint;
f:array[0..maxv] of boolean;
function max(a,b:longint):longint;
begin
if a>b then exit(a) else exit(b);
end;
procedure CompletePack(cost:longint);
var
i:longint;
begin
for i:=cost to v do
f[i]:=f[i] or f[i-cost];
end;
procedure ZeroOnePack(cost:longint);
var
i:longint;
begin
if cost=0 then exit;
for i:=v downto cost do
begin
f[i]:=f[i] or f[i-cost];
end;
end;
procedure main;
var
i,j,k,l,sum,p:longint;
begin
v:=v div 2;
for i:=1 to n do
begin
if i*a[i]>=v then
begin
CompletePack(i);
continue;
end;
sum:=1;
k:=0;
while (a[i]-sum+1>0) do begin inc(k); sum:=sum*2; end;
dec(k);
sum:=1;
for j:=0 to k-1 do
begin
ZeroOnePack(sum*i);
sum:=sum*2;
end;
ZeroOnePack((a[i]-sum+1)*i);
if f[v] then
begin
writeln('Can be divided.');
exit;
end;
end;
if not(f[v]) then writeln('Can''t be divided.')
else writeln('Can be divided.')
end;
begin
for i:=1 to n do read(a[i]);
j:=1;
while (a[1]+a[2]+a[3]+a[4]+a[5]+a[6]<>0) do
begin
writeln('Collection #',j,':');
fillchar(f,sizeof(f),false);
f[0]:=true;
v:=0;
for i:=1 to n do inc(v,i*a[i]);
if (v mod 2=1) then writeln('Can''t be divided.')
else
begin
main;
end;
writeln;
for i:=1 to n do read(a[i]);
inc(j);
end;
end.
POJ 2140(数学问题）
问n=a+a+1+a+2+...+a+k 的情况总数
n=(k+1)*a+(k+1)*k/2
=(k+1)(a+k/2)
n为整数，k+1为整数，(a+k/2)为整数，k为偶数，k+1为奇数
当n和k+1确定时，a为定值
故解为n的奇因子个数
Program P2140;
var
i,n,ans:longint;
begin
ans:=0;
read(n);
i:=1;
while (i<=n) do
begin
if (n mod i=0) then inc(ans);
inc(i,2);
end;
writeln(ans);
end.