easy

记录信息
评测状态 Accepted
题目 P1784 数字统计
递交时间 2014-08-03 12:07:12
代码语言 C++
评测机 VijosEx
消耗时间 15 ms
消耗内存 272 KiB
评测时间 2014-08-03 12:07:19
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 272 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 272 KiB, score = 10
Accepted, time = 15 ms, mem = 272 KiB, score = 100

#include <cstdlib>
#include <iostream>

using namespace std;

int gs(int a,int b)
{
int k=0;
while(a>0)
{
if(a%10==b)
k++;
a/=10;
}
return k;
}

int main(int argc,char *argv[])
{
int i,j,k=0,n,m;
cin>>n>>m;
j=2;
for(i=n;i<=m;i++)
{
k+=gs(i,j);
}
cout<<k<<endl;
// system("pause");
return 0;
}
//如果想把找2有多少个改一下,那就把j加到cin里.

Vijos题解：http://user.qzone.qq.com/616674147/2

Vijos 题解：http://hi.baidu.com/umule/item/2c997f8ed9600fdae596e017
有疑问请留言 共同进步

var n, m,k, i: longint;
begin
readln(n, m); k := 0;
for i := n to m do begin
n := i;
while n > 0 do begin if n mod 10 = 2 then Inc(k);n := n div 10; end;
end; writeln(k);
end.

Program two;
var l,r,i,j,num:integer;
s:string;

Begin
readln(l,r);
for i:=l to r do begin
str(i,s);
for j:=1 to length(s) do
if s[j]='2' then num:=num+1;
end;
writeln(num);
End.

#include <iostream>
using namespace std;

int main()
{
int x,y;
cin>>x>>y;
int count=0;
for(int i=x;i<=y;i++)
{
if (i/1000==2)
count++;
if (i%1000/100==2)
count++;
if (i%1000%100/10==2)
count++;
if (i%1000%100%10==2)
count++;
}
cout<<count<<endl;
return 0;
}

#include <iostream>
#include <stdio.h>
using namespace std;
int l,r,i,j,k,b,a,sum;
int main(){
cin>>l;
cin>>r;
sum=0;
for(i=l;i<=r;i++){
a=i;
do
{b=a%10;
if (b==2){
sum++;
}
a/=10;}
while(a!=0);
}
cout<<sum<<endl;
}
b表示取i(a)的每一位
如果是2就累加

刷水题神马的最有自信了O(∩_∩)O~

var l,r,x:integer;
begin
x:=0;
readln(l,r);
l:=l-1;
for l:=l+1 to r do
begin
if l mod 10=2 then x:=x+1;
if l div 10=2 then x:=x+1;
if l div 100=2 then x:=x+1;
if l div 1000=2 then x:=x+1;
end;
write(x);
end.
为啥只有30？

喵了个咪，居然是标准输入输出。那题干里还提输入输出文件。。
好久不刷题站了都忘了规矩了

├ 测试数据 01：答案正确... (156ms, 404KB)

├ 测试数据 02：答案正确... (31ms, 404KB)

├ 测试数据 03：答案正确... (47ms, 404KB)

├ 测试数据 04：答案正确... (0ms, 404KB)

├ 测试数据 05：答案正确... (16ms, 404KB)

├ 测试数据 06：答案正确... (0ms, 404KB)

├ 测试数据 07：答案正确... (31ms, 404KB)

├ 测试数据 08：答案正确... (16ms, 404KB)

├ 测试数据 09：答案正确... (47ms, 404KB)

├ 测试数据 10：答案正确... (47ms, 404KB)

这题本人上传的！

首先，我们可以让一个变量从最小值循环制最大值，然后把每个循环到的数字转换为字符串，再用一个变量循环，判断每一位是否为2，是就加1！

很水的

var

s,n,q,e,l:integer;

w:string;

begin

read(s,n);

for q:=s to n do begin

str(q,w);

for e:=1 to length(w)do if w[e]='2' then l:=l+1;

end;

writeln(l);

end.

AC31 一星纪念

├ 测试数据 01：答案正确... (0ms, 676KB)

├ 测试数据 02：答案正确... (15ms, 676KB)

├ 测试数据 03：答案正确... (0ms, 676KB)

├ 测试数据 04：答案正确... (15ms, 676KB)

├ 测试数据 05：答案正确... (0ms, 676KB)

├ 测试数据 06：答案正确... (0ms, 676KB)

├ 测试数据 07：答案正确... (15ms, 676KB)

├ 测试数据 08：答案正确... (15ms, 676KB)

├ 测试数据 09：答案正确... (15ms, 676KB)

├ 测试数据 10：答案正确... (15ms, 676KB)

这题......真是水额，5分钟AC

├ 测试数据 01：答案正确... (13ms, 672KB)

├ 测试数据 02：答案正确... (14ms, 672KB)

├ 测试数据 03：答案正确... (12ms, 672KB)

├ 测试数据 04：答案正确... (31ms, 672KB)

├ 测试数据 05：答案正确... (13ms, 672KB)

├ 测试数据 06：答案正确... (0ms, 672KB)

├ 测试数据 07：答案正确... (0ms, 672KB)

├ 测试数据 08：答案正确... (15ms, 672KB)

├ 测试数据 09：答案正确... (0ms, 672KB)

├ 测试数据 10：答案正确... (0ms, 672KB)

---|---|---|---|---|---|---|---|-

Accepted / 100 / 101ms / 672KB

农夫山泉有点甜

第二个AC

var a,b,t,i,s:longint;
begin
read(a,b);
for i:=a to b do
begin
t:=i;
while t>0 do
begin
if t mod 10=2 then inc(s);
t:=t div 10;
end;
end;
write(s);
end.

一组要把数组开大。。。。开10000一直过不了。。。。因为数组10000的值是你输入的第二值的值。。。。所以一定多开一个
```c
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
int m,n;
int i,j;
int k;
int sh[10001]= {0};
for(i=1; i < 10000; i++)
{
j=i;
k=0;
while(j != 0)
{
if(j%10 == 2)
{
if(k == 0){sh[i]=1;k = 1;}
else sh[i]++;
}
j /= 10;
}
// printf("i = %d sh[i] = %d\n",i,sh[i]);
}
scanf("%d%d",&m,&n);
int sum=0;
// printf("%d",sh[10000]);
for(i=m; i <= n; i++)
{
if(sh[i] > 0) sum += sh[i];
}
printf("%d\n",sum);

return 0;
}

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真的太水
var
l,r,i,j,n,ans:longint;
s:string;
begin
readln(l,r);
for i:=l to r do begin
str(i,s);
n:=length(s);
for j:=1 to n do begin
if s[j]='2' then inc(ans);
end;
end;
writeln(ans);
end.