335 条题解
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0
rock lee LV 7 @ 2010-07-23 21:29:35
一次ac
新手过的第一道难度三的题！good！
规规矩矩的做法，没有压位，没有那么复杂，10进制普通做法就行
string要换
program p1040;
var
l1,l2,l3,i,j,k,l:longint;
a,b:array[0..999999]of longint;
s1,s2:ansistring;
begin
readln(s1);readln(s2);
l1:=length(s1);l2:=length(s2);
for i:=0 to l1-1 do
a[i]:=ord(s1[l1-i])-ord('0');
for i:=0 to l2-1 do b[i]:=ord(s2[l2-i])-ord('0');
for i:=l1-1 downto 0 do
begin
for j:=l2-1 downto 1 do
a:=a+a[i]*b[j];
a[i]:=a[i]*b[0];
end;
k:=l1+l2-1;
for i:=0 to k-1 do
begin
a:=a[i] div 10 +a;
a[i]:=a[i] mod 10;
end;
if a[k]>0 then k:=k+1;
for i:=k-1 downto 0 do write(a[i]);
end.
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 431ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：431ms -
0
var
s1,s2:ANSIstring;
a,b:array[1..10000]of longint;
c:array[1..1000000]of longint;
i,j:longint;
begin
readln(s1);
readln(s2);
fillchar(a,sizeof(a),0);
fillchar(b,sizeof(b),0);
fillchar(c,sizeof(c),0);
for i:=1 to length(s1)do
a[length(s1)-i+1]:=ord(s1[i])-48;
for i:=1 to length(s2)do
b[length(s2)-i+1]:=ord(s2[i])-48;
for i:=1 to length(s1)do begin
for j:=1 to length(s2)do begin
c[j+i-1]:=c[j+i-1]+a[i]*b[j];
end;
end;
j:=0;
repeat
inc(j);
if c[j]>=10
then begin
c[j+1]:=c[j+1]+c[j] div 10;
c[j]:=c[j] mod 10;
end;
until c[j+1]=0;
for i:=j downto 1 do
write(c[i]);
writeln;
end.
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：运行超时...
---|---|---|---|---|---|---|---|-
Unaccepted 有效得分：75 有效耗时：0ms
随能帮我看看为什么超时
或怎样才不超时！ -
0
program p1040_1;
type arr=record
s:array[0..5000]of int64;
len:longint;
end;
var
x,i,j,l:longint;
str:ansistring;
a,b:arr;function mul(a,b:arr):arr;
var
c:arr;
i,j:longint;
begin
fillchar(c,sizeof(c),0);
for i:=1 to a.len do
for j:=1 to b.len do
begin
inc(c.s,a.s[i]*b.s[j]);
inc(c.s,c.s div 100000000);
c.s:=c.s mod 100000000;
end;
c.len:=a.len+b.len+1;
while (c.s[c.len]=0)and(c.len>1) do dec(c.len);
mul:=c;
end;
begin
readln(str);
l:=length(str);
a.len:=l div 8;
for i:=1 to a.len do
val(copy(str,l-i*8+1,8),a.s[i]);
if l mod 80 then begin inc(a.len);val(copy(str,1,l mod 8),a.s[a.len]); end;
readln(str);
l:=length(str);
b.len:=l div 8;
for i:=1 to b.len do
val(copy(str,l-i*8+1,8),b.s[i]);
if l mod 80 then begin inc(b.len);val(copy(str,1,l mod 8),b.s); end;
a:=mul(a,b);
write(a.s[a.len]);
for i:=a.len-1 downto 1 do
begin
x:=10000000;
if a.s[i]=0 then
for j:=1 to 8 do
write(0);
while a.s[i]0 do
begin
if a.s[i] div x=0 then begin write(0);x:=x div 10; end
else begin write(a.s[i]);a.s[i]:=0 end;
end;
end;
end.压了8位了 为什么还是没有秒杀 哎
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f531609175 @ 2013-08-01 20:36:36
用了过程会浪费一点点时间
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0
压位
program p1040;
var a,b,c:array[0..20010] of longint;
k1,k2,k,i,j,n1,n2,jsq:longint;
x:char;
st:ansistring;
begin
readln(st);
n1:=length(st);
k1:=7-(n1 mod 4);
jsq:=0;
while jsq='0') and (x0) do k:=k-1;
if k=0 then writeln(0) else
begin
write(c[k]);
for i:=k-1 downto 1 do
begin
if c[i]>999 then write(c[i])
else if c[i]>99 then write('0',c[i])
else if c[i]>9 then write('00',c[i])
else write('000',c[i]);
end;
writeln;
end;
end. -
0
#include"stdio.h"
#include"string.h"
int main()
{
char m[15000],n[15000];
static long int im[3000],in[3000];
long int sum[100000]={0};
int mlen,nlen;
int i,ii=0,j,jj=0,mk=0,nk=0,sk=0;
int x;
int t=0;
int flag=0;
scanf("%s",m);
getchar();
scanf("%s",n);
mlen=strlen(m);
nlen=strlen(n);for(i=0;i
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0
#include
#include
using namespace std;
int a[40001],b[40001],s[40001];
int la,lb,ls,i,ja,ka,jb,kb,j,k;
string sa,sb;
int main(){
cin>>sa;
ka=sa.length();
for(i=1;i>2;
a[ja]=a[ja]*10+sa-48;
}
la=(ka+3)>>2;
cin>>sb;
kb=sb.length();
for(i=1;i>2;
b[jb]=b[jb]*10+sb-48;
}
lb=(kb+3)>>2;
ls=0;
for(i=1;i -
0
var
a,b:array[0..40001] of longint;
l1,l2,len,i,k:longint;
str:ansistring;
procedure multiply;
var i,j:integer;
begin
for i:=l1-1 downto 0 do
begin
for j:=l2-1 downto 1 do
inc(a,a[i]*b[j]);
a[i]:=a[i]*b[0];
end;
for i:=0 to k-1 do
begin
inc(a,a[i] div 10);
a[i]:=a[i] mod 10;
end;
if a[k]0 then inc(k,1);
end;
begin
readln(str);
len:=length(str);
l1:=len;
for i:=0 to len-1 do
a[len-i-1]:=ord(str)-ord('0');readln(str);
len:=length(str);
l2:=len;
for i:=0 to len-1 do
b[len-i-1]:=ord(str)-ord('0');k:=l1+l2-1;
multiply;
for i:=k-1 downto 0 do
write(a[i]);
end.
谁说用万进制了？
我虽然耗时384MS 但用的是十进制的！
数据开大点 不是过不了！ -
0
var
m,n:ansistring;
a,b,c,kk:array[1..50000000] of longint;
len1,len2,len3,i,j,t:longint;begin
kk[1]:=1;
read(m);readln;readln(n);
len1:=length(m);
len2:=length(n);
for i:=1 to len1 do a[i]:=ord(m[len1-i+1])-48;
for i:=1 to len2 do b[i]:=ord(n[len2-i+1])-48;begin
for i:=1 to len1 do
for j:=1 to len2 do
begin
c:=c+a[i]*b[j];
end;
end;len1:=len2+len1;
for i:=1 to len1 do if c[i]>=10 then
begin
c:=c[i] div 10 +c;
c[i]:=c[i] mod 10;
end;for i:=1 to len1 do if c[i]0 then t:=i;
for i:=t downto 1 do write(c[i]);
end.太烦了。。。。又过老。。太简单老。。。。离我得ioi满分的日子已经几年老。。。。好想念那些在希腊参赛的日子。。。。好想那个老师。。。。
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2pac @ 2014-03-18 19:38:06
fuck your mother 编译错误
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0
万进制
数组开不够 说我超时。。。编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 353ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：353msprogram ex;
type arr=array[0..5000]of int64;
var i,j:longint;
s1,s2:ansistring;
a,b,c:arr;procedure init;
var i,j:longint;
begin
readln(s1);
while (s1[1]='0')and(length(s1)>1) do delete(s1,1,1);
readln(s2);
while (s2[1]='0')and(length(s2)>1) do delete(s2,1,1);
a[0]:=length(s1) div 4 +ord(length(s1) mod 40);
for i:=1 to a[0]-1 do
begin
val(copy(s1,length(s1)-3,4),a[i]);
delete(s1,length(s1)-3,4);
end;
val(s1,a[a[0]]);
b[0]:=length(s2) div 4 +ord(length(s2) mod 40);
for i:=1 to b[0]-1 do
begin
val(copy(s2,length(s2)-3,4),b[i]);
delete(s2,length(s2)-3,4);
end;
val(s2,b[b[0]]);
end;procedure mul;
var i,j:longint;
begin
fillchar(c,sizeof(c),0);
c[0]:=a[0]+b[0]-1;
for i:=1 to a[0] do
for j:=1 to b[0] do
inc(c,a[i]*b[j]);
for i:=1 to c[0] do
begin
inc(c,c[i] div 10000);
c[i]:=c[i] mod 10000;
end;
while c[c[0]+1]>0 do
begin
inc(c[0]);
inc(c[c[0]+1],c[c[0]] div 10000);
c[c[0]]:=c[c[0]] mod 10000;
end;
while (c[c[0]]=0)and(c[0]>1) do dec(c[0]);
end;procedure print;
var i,j:longint;
ans:ansistring;
begin
write(c[c[0]]);
for i:=c[0]-1 downto 1 do
begin
if c[i] -
0
var n,m,x,y:ansistring;
q,w,e,i,j,l,k,z1,z2:longint;
a,b,c:array[1..100000]of int64;
procedure cheng;
var j,i,r,u:longint;
begin
u:=0;
for i:=1 to q do
for j:=1 to w do begin
c:=a[i]*b[j]+c;
c:=c+cdiv 10000;
c:=cmod 10000;
if c[q+w]0 then u:=1;
end;
for i:=q+w-1+u downto 1 do
begin
if iq+w-1+u then
if c[i] -
0
program gjdcf;
var a,b,c,d:array[1..50] of longint;
i,j,k,p:longint;begin
fillchar(a,sizeof(a),100);
fillchar(b,sizeof(b),100);randomize;
for i:=20 downto 1 do
begin
a[i]:=random(9)+1;
write(a[i]);
end;
write('*');
writeln;
for i:=20 downto 1 do
begin
b[i]:=random(9)+1;
write(b[i]);
end;
write('=');
writeln;for j:=20 downto 1 do
c[j]:=a[j];for i:=1 to 20 do
begina:=a+(a[i]*b[i])div 10;
a[i]:=a[i]*b[i];while a[i] >=10 do
begin
a[i]:=a[i]mod 10;
k:=a[i];
inc(i);
a[i]:=a[i]+(k div 10);
end;while
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0
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 259ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：259ms我果然善于把程序写得很长很长……
const filename='p1040';
var
a,b:array[1..2500]of longint;
c:array[1..5000]of longint;
s:ansistring;
i,j,l,l2,x,k:longint;
begin
assign(input,filename+'.in');reset(input);
assign(output,filename+'.out');rewrite(output);
readln(s);
l:=length(s)div 4;
if length(s)mod 40then
begin
inc(l);
for i:=l downto 2 do
begin
val(copy(s,length(s)-3,4),a[l-i+1]);
delete(s,length(s)-3,4);
end;
val(s,a[l]);
end
else
for i:=l downto 1 do
begin
val(copy(s,length(s)-3,4),a[l-i+1]);
delete(s,length(s)-3,4);
end;
// for i:=1 to l do writeln(a[i]); writeln;
///////////////
readln(s);
l2:=length(s)div 4;
if length(s)mod 40then
begin
inc(l2);
for i:=l2 downto 2 do
begin
val(copy(s,length(s)-3,4),b[l2-i+1]);
delete(s,length(s)-3,4);
end;
val(s,b[l2]);
end
else
for i:=l2 downto 1 do
begin
val(copy(s,length(s)-3,4),b[l2-i+1]);
delete(s,length(s)-3,4);
end;
//for i:=1 to l2 do writeln(b[i]);
for i:=1 to l do
for j:=1 to l2 do
begin
c:=a[i]*b[j]+c;
if c>10000 then
begin
c:=c[j+i-1]div 10000+c;
c:=cmod 10000;
end
end;
l:=l+l2;
while c[l]=0do dec(l);write(c[l]);
for i:=l-1 downto 1 do
begin
k:=c[i];
for j:=3 downto 1 do
begin
if k div 10=0then write(0);
k:=k div 10;
end;
write(c[i]);
end;
writeln;
close(input);close(output);
end.友情提供二数据（自己编的）：
in
1564665465
775795268out
1213860063750019620in
443
1432531out
634611233这个程序在桌面上留个.exe以后用来代替calc.exe计算高精乘法……
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0
song19931218 LV 9 @ 2009-10-10 19:21:19
压四位，秒杀
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broken-boat LV 7 @ 2009-10-06 09:36:10
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：0ms难得一次AC撒花庆祝一下,,不过MS我的程序时最长的囧
10^8进制program s1040;
var
a,b,c,d,ans,final:array [1..30000] of qword;
s:ansistring;
i,j,nb,nc,m,n:longint;
procedure doe;
var i,j:longint;
begin
for i:=1 to nb do
begin
fillchar(d,sizeof(d),0);
for j:=1 to nc do
d[j+i-1]:=c[j]*b[i];
for j:=1 to nc+i-1 do
begin
ans[j]:=d[j]+ans[j];
if ans[j]>=100000000 then
begin
ans[j+1]:=ans[j+1]+ans[j] div 100000000;
ans[j]:=ans[j] mod 100000000;
end;
end;
end;
end;
{main}//其实就是读入输出= =
begin
readln(s);
fillchar(a,sizeof(a),0);
for i:=1 to length(s) do
a[i]:=ord(s[length(s)-i+1])-48;
if length(s) mod 8 =0 then nb:=length(s) div 8 else
nb:=length(s) div 8+1;
for i:=1 to nb do
begin
m:=0;
for j:=8 downto 1 do
m:=m*10+a[(i-1)*8+j];
b[i]:=m;
end;
readln(s);
fillchar(a,sizeof(a),0);
for i:=1 to length(s) do
a[i]:=ord(s[length(s)-i+1])-48;
if length(s) mod 8 =0 then nc:=length(s) div 8 else
nc:=length(s) div 8+1;
for i:=1 to nc do
begin
m:=0;
for j:=8 downto 1 do
m:=m*10+a[(i-1)*8+j];
c[i]:=m;
end;
fillchar(a,sizeof(a),0);
//读数直接复制了一下= =写个过程MS也不错doe; //操作
n:=nb+nc+1;
while (ans[n]=0) and (n>=2) do dec(n);
for i:=1 to n do
for j:=1 to 8 do
begin
final[(i-1)*8+j]:=ans[i] mod 10;
ans[i]:=ans[i] div 10;
end;
n:=n*8;
while (final[n]=0) and (n>=2) do dec(n);
for i:=n downto 1 do
write(final[i]);
end.
rock lee
LV 7
@ 2010-07-23 21:29:35
