各位同学呀，作者道题千万要仔细啊，最后一组数据比10000大，开数组时多开一些否则很惨的。20次AC，悲哀啊。

两次AC

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├ 测试数据 02：答案正确... 0ms

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Accepted 有效得分：100 有效耗时：0ms

1E9进制高精度

输入输出有点麻烦

不知道为什么1E9比1E1快这么多？

program vijos_1040; { 高精度乘法 }

const

maxn=20001;

base=100;

logBase=2;

type

baseType=qword;

highInt=record

da:array[1..maxn] of baseType;

len:integer;

end;

highPoint=^highInt;

var

a,b,c:highInt;

procedure empty(var a:highInt);

begin

with a do

begin

len:=1;

fillchar(da,sizeof(da),0);

end;

end;

procedure val(var s:ansistring;var a:highInt);

var

ch:char;

i,j,k:integer;

begin

with a do

begin

while length(s) mod logBase0 do s:='0'+s;

j:=0;

k:=length(s);

while k>0 do

begin

inc(j);

for i:=k-logBase+1 to k do

da[j]:=da[j]*10+ord(s[i])-48;

dec(k,logBase);

end;

len:=j;

end;

end;

procedure init;

var

s:ansistring;

po,len:integer;

begin

empty(a);

empty(b);

empty(c);

readln(s);

val(s,a);

readln(s);

val(s,b);

end;

procedure highMul(const a,b:highInt;var c:highInt);

var

i,j,l:integer;

begin

empty(c);

for i:=1 to a.len do

for j:=1 to b.len do

begin

inc(c.da,a.da[i]*b.da[j]);

if c.da>=base then

begin

inc(c.da,c.da div base);

c.da:=c.da mod base;

end

end;

l:=a.len+b.len+1;

while (l>0) and (c.da[l]=0) do dec(l);

c.len:=l;

end;

procedure print(const a:highInt);

var

i,j:integer;

begin

with a do

for i:=len downto 1 do

begin

if ilen then

for j:=1 to logBase-trunc(ln(da[i])/ln(10))-1 do

write(0);

write(da[i]);

end;

writeln;

end;

begin

init;

highMul(a,b,c);

print(c);

end.

program p1040;

var s:ansistring;

a,b,ans:array[1..10000]of longint;

m,n,i,j,k,l,r,tt:longint;

procedure print(i:longint);

var j,t:longint;

begin

if (i=k)or(ans[i] div 10000) then write(ans[i])

else

begin

t:=ans[i];

for j:=1 to 4 do

begin

t:=t div 10;

if t=0 then break;

end;

t:=4-j;

for j:=1 to t do write('0');

write(ans[i]);

end;

end;

begin

readln(s);

k:=length(s);

m:=0;

i:=1;

while i

楼下那位的能过吗？

program P_1027;

type

xx=array[0..520]of longint;

var

a,b,c:xx;

s:string;

i:longint;

function cheng(a,b:xx):xx;

var

i,j:longint;

c:xx;

begin

fillchar(c,sizeof(c),0);

c[0]:=a[0]+b[0];

for i:=1 to a[0] do

for j:=1 to b[0] do

begin

c:=a[i]*b[j]+c;

c:=c div 10 +c;

c:=c mod 10;

end;

while (c[0]>1)and(c[c[0]]=0) do dec(c[0]);

cheng:=c;

end;

begin

readln(s);

a[0]:=length(s);

for i:=1 to a[0] do

a[i]:=ord(s[a[0]+1-i])-48;

readln(s);

b[0]:=length(s);

for i:=1 to b[0] do

b[i]:=ord(s[b[0]+1-i])-48;

c:=cheng(a,b);

for i:=c[0] downto 1 do

write(c[i]);

writeln;

end.

var

s1,s2,s3,s:ansistring;

i,j,k1,k2,l1,l2,t1,t2,m,x,y,z,w:longint;

a,b,c:array[1..5000]of longint;

begin

readln(s1);

l1:=length(s1);

k1:=l1;

t1:=1;

while k1>=4 do

begin

s:=copy(s1,k1-3,4);

val(s,m);

a[t1]:=m;

inc(t1);

dec(k1,4);

end;

if l1 mod 40 then begin

s:=copy(s1,1,l1 mod 4);

val(s,m);

a[t1]:=m;

end

else dec(t1);

readln(s2);

l2:=length(s2);

k2:=l2;

t2:=1;

while k2>=4 do

begin

s:=copy(s2,k2-3,4);

val(s,m);

b[t2]:=m;

inc(t2);

dec(k2,4);

end;

if l2 mod 40 then begin

s:=copy(s2,1,l2 mod 4);

val(s,m);

b[t2]:=m;

end

else dec(t2);

for i:=1 to t1 do

for j:=1 to t2 do

begin

x:=a[i]*b[j];

y:=x div 10000;

z:=x mod 10000;

w:=i+j-1;

c[w]:=c[w]+z;

c[w+1]:=c[w+1]+c[w]div 10000+y;

c[w]:=c[w] mod 10000;

end;

m:=t1+t2;

if c[m]=0 then dec(m);

write(c[m]);

for i:=m-1 downto 1 do

begin

if (c[i]>=1000)and(c[i]=100)and(c[i]=10)and(c[i]=0)and(c[i]

普通乘法，只是要用ansistring！水题！

program chengfa;

var str1,str2:string;

n1,n2,i,j,n,w,p,q:integer;

a,b:array[1..256] of integer;

c:array[1..1000] of integer;

begin

readln(str1);readln(str2);

n1:=length(str1);

n2:=length(str2);

for i:=n1 downto 1 do val(str1[i],a[n1-i+1]);

for i:=n2 downto 1 do val(str2[i],b[n2-i+1]);

for i:=1 to (n1+n2) do c[i]:=0;

for i:=1 to n2 do

for j:=1 to n1 do

begin

w:=i+j-1;q:=c[w];

c[w]:=(q+(a[j]*b[i]) mod 10) mod 10;

c[w+1]:=c[w+1]+(q+(a[j]*b[i]) mod 10) div 10+ (a[j]*b[i]) div 10;

end;

for i:=(n1+n2) downto 1 do

if (c[i]0) then

begin

n:=i;break;

end;

for i:=n downto 1 do write(c[i]);

end.

var

s1,s2,s3:ansistring;

a,b,c,d:array[0..5000]of longint;

i,j,k,l1,l2,la,lb,ka,kb,temp:longint;

begin

readln(s1);

readln(s2);

if length(s1)=0)and(c[j]=10)and(c[j]=100)and(c[j]

TMD ~~~~~~~~`竟然会 超时~~~~~~~\ 怎么可能???????

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... 462ms

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Accepted 有效得分：100 有效耗时：462ms

program p1040;

var

a,b,c:array[1..20001] of longint;

i,j,k,t,p,q:longint;

ch:char;

begin

p:=0;

while not eoln do begin

read(ch);

inc(p);

a[p]:=ord(ch)-48;

end;

readln;

q:=0;

while not eoln do begin

read(ch);

inc(q);

b[q]:=ord(ch)-48;

end;

for i:=1 to p div 2 do begin

k:=a[i];

a[i]:=a[p+1-i];

a[p+1-i]:=k;

end;

for i:=1 to q div 2 do begin

k:=b[i];

b[i]:=b[q+1-i];

b[q+1-i]:=k;

end;

fillchar(c,sizeof(c),0);

for i:=1 to p do

for j:=1 to q do

c:=c+a[i]*b[j];

for i:=1 to p+q do begin

c:=c+c[i] div 10;

c[i]:=c[i] mod 10;

end;

k:=20001;

while c[k]=0 do dec(k);

for i:=k downto 1 do

write(c[i]);

end.

第一次写压位高精度。感觉不错

0MS

普通一位高精可以!

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... 541ms

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Accepted 有效得分：100 有效耗时：541ms

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

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Accepted 有效得分：100 有效耗时：0ms

第一次写压位高精，居然卡了我一下午。。。= =

居然错在读入上了。。囧。。。

var i,j,k,l1,l2,m,g,n,c,x,z:longint;

a,b,y,jia:array[1..1000] of longint;

s1,s2:ansistring;

begin

readln(s1); l1:=length(s1); n:=500;

readln(s2); l2:=length(s2); z:=0;

for i:=l1 downto 1 do val(s1[i],a[n+i-l1],z);

for i:=l2 downto 1 do val(s2[i],b[n+i-l2],z);

for i:=n downto n-l2+1 do

begin

g:=0;

for j:=500 downto 1 do

begin

c:=a[j]*b[i]+g;

y[j]:=c mod 10;

g:=c div 10;

end;

g:=0;

for j:=500 downto 1 do

begin

c:=y[j]+jia[j]+g;

jia[j]:=c mod 10;

g:=c div 10;

end;

fillchar(y,sizeof(y),0);

z:=z+1;

end;

m:=1;

while jia[m]=0 do inc(m);

for i:=m to 500 do write(jia[i]);

end.

哪个地方错了~~5555

要压位.

交了5次才AC 太不容易了，开始数组里存一位数，最后一个数据超时，后来改成4位数…………………………

program p1040;

var a,b:array[-5..10000] of longint;

c:array[-5..100000000] of longint;

ch:ansistring;

l1,l2,m,k:longint;

procedure goin;

var l,min,i,j:longint; s:string;

begin

readln(ch); s:='';min:=10000;

l1:=length(ch);

for i:=l1 downto 1 do

begin

if length(s)=100 then write('0',c[m]) else

if c[m]>=10 then write('00',c[m]) else

if c[m]>=1 then write('000',c[m]) else

write('0000');

end;

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：答案正确... 0ms

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Accepted 有效得分：100 有效耗时：0ms

10000位压位，一次AC，秒杀……

#include "stdio.h"

#include "string.h"

int main()

{ char a[10005],b[10005];

int x[10005]={0},y[10005]={0},s[20015]={0},i,j,n,m,d=0,t=0,t1=0,len;

scanf("%s%s",a,b);

n=strlen(a);

m=strlen(b);

for(i=n-4;i>=0;i-=4)

x[t++]=(a[i]-'0')*1000+(a-'0')*100+(a-'0')*10+(a-'0')*1;

if(n-t*4==3)

x[t]=(a[0]-'0')*100+(a[1]-'0')*10+(a[2]-'0')*1;

else

if(n-t*4==2)

x[t]=(a[0]-'0')*10+(a[1]-'0')*1;

else

if(n-t*4==1)

x[t]=a[0]-'0';

for(j=m-4;j>=0;j-=4)

y[t1++]=(b[j]-'0')*1000+(b[j+1]-'0')*100+(b[j+2]-'0')*10+(b[j+3]-'0')*1;

if(m-t*4==3)

y[t1]=(b[0]-'0')*100+(b[1]-'0')*10+(b[2]-'0')*1;

else

if(m-t*4==2)

y[t1]=(b[0]-'0')*10+(b[1]-'0')*1;

else

if(m-t*4==1)

y[t1]=b[0]-'0';

for(i=0;i=0;i--,t1++)

if(b[i]!=0)

break;

for(i=len;i>=0;i--)

{

if(i==len)

printf("%d",s[i]);

else

printf("%.4d",s[i]);

}

return 0;

}

type arra=array[-3..10000]of longint;

var a,b,c:arra;

m,n,next,l1,l2,l3:longint;

procedure init(var m:arra;var lo:longint);

var k:array[-4..10000]of char;

q:array[-3..10000]of longint;

l,i,j,z,p:longint;

s:string;

begin

l:=0;

fillchar(m,sizeof(m),0);

while not(eoln) do

begin

inc(l);

read(k[l]);

end;

for i:=-4 to 0 do k[i]:='0';

for i:=1 to (l div 3)+1 do

begin

s:='000';

z:=1;

for j:=l-i*3+1 to l-(i-1)*3 do

begin

s[z]:=k[j];

inc(z);

end;

val(s,q[i]);

end;

readln;

lo:=(l div 3)+1;

m:=q;

end;

procedure multiplied(a,b:arra;l1,l2:longint);

var i,j:longint;

s:string;

begin

l3:=l1+l2;

next:=0;

fillchar(c,sizeof(c),0);

for i:=1 to l1 do

for j:=1 to l2 do

begin

c:=a[i]*b[j]+c;

if c>=1000 then

begin

c:=c+c div 1000;

c:=c mod 1000;

if i+j>l3 then inc(l3);

end;

end;

while c[l3]=0 do dec(l3);

write(c[l3]);

for i:=l3-1 downto 1 do

begin

str(c[i],s);

while length(s)

高精 压位啊