题解 - Vijos

0

Daqu LV 5 @ 2017-05-23 20:11:16

用c++写的，思路是用一个递归函数处理每个数，然后用for循环处理全部的数

#include <iostream>

long int count(long int temp, long int x);

int main()
{
    long int n, x, result=0;
    std::cin >> n >> x;
    for(long int i=1;i<=n;i++)
    {
        result+=count(i, x);
    }

    std::cout << result << std::endl;

    return 0;
}

long int count(long int temp, long int x)
{
    if(temp/10 < 1)
    {
        if(temp == x)
        {
            return 1;
        }else
        {
            return 0;
        }
    }else
    {
        if(temp%10 == x)
        {
            return 1 + count(temp/10, x);
        }else
        {
            return 0 + count(temp/10, x);
        }
    }
}

0

ccl66 LV 7 @ 2017-05-20 21:13:30

#include<stdio.h>
#include<string.h>
char c[1100000];
int main()
{
int n,x,ans=0;

while(~scanf("%d%d",&n,&x)){
ans=0;
for(int i=1;i<=n;i++)
{
int j=i;
while(j>=10)
{
if(j%10==x){
ans++;
}
j/=10;
}
if(j==x)ans++;
}
printf("%d\n",ans);
}
return 0;
}

0

xuyifeng LV 8 @ 2017-05-13 14:06:21

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>

using namespace std;

int main(){
int n, x;
cin >> n >> x;
int ans = 0;
for(int i = 1; i <= n; i++){
int num = i;
while(num){
if(num % 10 == x) ans++;
num /= 10;
}
}
printf("%d", ans);
return 0;
}

0

w568w LV 8 @ 2017-05-08 13:09:37

一种奇葩的思路，用字符串
（我才不在乎效率呢）

#include <stdio.h>
int main()
{
    long n,times=0;
    char buf[7],x;
    scanf("%ld %c",&n,&x);
    for(long i=1;i<=n;i++)
    {
        int len=sprintf(buf,"%ld",i);
        for(int j=0;j<len;j++)
            if(buf[j]==x)
                times++;
    }
    printf("%ld",times);
}

0

15211160230 LV 7 @ 2017-05-06 21:08:36

import java.util.Scanner;
public class Main{
    public static void main(String args[]){
        Scanner in=new Scanner(System.in);
        int n,x,count=0,t;
        String now,index;
        n=in.nextInt();
        x=in.nextInt();
        index=String.valueOf(x);
        for(int i=1;i<=n;++i){
            now=String.valueOf(i);
            t=0;
            while(true){
                t=now.indexOf(index,t);
                if(t==-1)   break;
                count++;
                t++;
            }
        }
        System.out.println(count);
        in.close();
    }
}

0

lll5810 LV 8 @ 2017-05-06 17:43:12

#include <iostream>
#include <cstdio>
#include <string.h>
#include <cmath>
using namespace std;
int main(){
long long n,temp;
int x;
int cnt = 0;
cin>>n>>x;
// cout<<n<<" "<<x<<endl;
for(long long i=1; i<=n ;i++){
// cout<<i<<endl;
temp = i;
while(temp){
if(temp%10==x)
cnt++;
temp /= 10;
}
}
cout<<cnt<<endl;
return 0;
}

0

村雨sup LV 4 @ 2017-04-25 19:20:26

#include<stdio.h>
int main()
{
int sum=0,n,i,j,x;
scanf("%d %d",&n,&x);
for(i=1;i<=n;i++)
{
j=i;
while(j>0)
{
if(j%10==x)
sum++;
j=j/10;
}
}
printf("%d",sum);

}

0

Dinghow LV 6 @ 2017-04-14 12:37:55

#include<iostream>
using namespace std;

int main(){
    int n,x,temp,count = 0;
    cin>>n>>x;
    for(int j = 1 ; j <= n ; j++){
        int i = j;
        while(1){
            temp = i%10;
            if(x == temp)
                count++;
            if(i/10 == 0)
                break;
            else
                i/=10;
        }
    }
    cout<<count;
    return 0;
}

0

王润锋 LV 7 @ 2017-03-31 11:54:10

#include <stdio.h>
int main()
{
long n,i;
int x;
scanf("%d%d",&n,&x);
int a[10];
for(i=1;i<=9;i++) a[i]=0;
for(i=1;i<=n;i++)
{
long j=i;
while(j>0)
{
a[j%10]++;
j=j/10;
}

}
printf("%d",a[x]);
}

2714040588 @ 2017-11-06 21:01:13

厉害了，没想到可以用数组
无言 @ 2017-11-28 22:52:38

int a[10];要在scanf("%d%d",&n,&x);之前
lijie201602 @ 2018-01-21 14:24:41

@无言: 为什么呢，定义数组和n、x的输入有关么？

0

Creeper_Zhang LV 6 @ 2017-02-22 13:21:07

#include <stdio.h>
#include <stdlib.h>
int main()
{
int n,x,a,i,sum=0;
scanf("%d %d",&n,&x);
for(i=1;i<=n;i++)
{
a=i;
while(a>0)
{

if(a%10==x)
sum++;
a=a/10;
}
}
printf("%d",sum);
return 0;
}

0

Peisir LV 8 @ 2017-02-21 20:54:41

从1数到n，对于每一个数分离累加

#include <iostream>
using namespace std;
int num[10];

void f(int n)
{
    while(n != 0){
        num[n % 10]++;
        n /= 10;
    }
    return ; 
}
int main(void)
{
    int n,t,i;
    cin>>n>>t;

    for(i = 1;i <= n;i++)
    {
        f(i);
    }

    cout<<num[t];
    return 0;
}

0

xanarry LV 5 @ 2017-02-14 16:43:45

#include <stdio.h>
int main(int argc, char const *argv[])
{
int n,x,j,k=0;
scanf("%d%d",&n,&x);
for(int i=1;i<=n;i++)
{
j=i;
while(j!=0)
{
if(x==j%10)
{
k+=1;
}
j=j/10;
}
}
printf("%d\n",k );
return 0;
}

0

TigerHou LV 5 @ 2017-02-06 21:07:06

#include<iostream>
using namespace std;
int main()
{
    int n, x;
    cin >> n >> x;
    int sum = 0;
    for (int i = 1; i <= n; i++)
    {
        for (int j = i; j; j /= 10)
        {
            sum += ((j % 10) == x);
        }
    }
    cout << sum << endl;
    system("pause");
    return 0;
}

个人认为，c++的话还是数这种方法比较简单

0

conan1983105 LV 6 @ 2017-01-13 20:23:43

#include<cstdio>
#define maxn 1000005
int sh[maxn][10]={0};
int main()
{

int j,m;
for(int i=1;i<=maxn;i++)
{
j=i;
while(j!=0)
{
m = j%10;
sh[i][m]++;
j=j/10;
}
}
for(int i=1;i<=1000000;i++)
for(int j=0;j<10;j++)
sh[i][j] = sh[i-1][j]+sh[i][j];

int n,x;
scanf("%d%d",&n,&x);
printf("%d\n",sh[n][x]);
return 0 ;
}

0

gg_c LV 7 @ 2017-01-03 00:12:40

水水水，写一遍，编译一遍，提交一遍。直接先要打印的数用二维数组存起来就好，用的时候直接打印就成。
```c++
#include<cstdio>
#define maxn 1000005
int sh[maxn][10]={0};
int main()
{

int j,m;
for(int i=1;i<=maxn;i++)
{
j=i;
while(j!=0)
{
m = j%10;
sh[i][m]++;
j=j/10;
}
}
for(int i=1;i<=1000000;i++)
for(int j=0;j<10;j++)
sh[i][j] = sh[i-1][j]+sh[i][j];

int n,x;
scanf("%d%d",&n,&x);
printf("%d\n",sh[n][x]);
return 0 ;
}

miaojie4321 @ 2017-02-26 13:32:43

我想问一下为什么不能while（i）
一定要先j=i
我用上面编出的是死循环

0

RedContritio LV 4 @ 2016-12-22 17:33:52

//最简短实现
#include<stdio.h>
int i,k,n,p,x;
int main()
{
scanf("%d%d",&n,&x);
for(i=1;i<=n;i++)for(k=i;k;k/=10)p+=((k%10)==x);
printf("%d",p);
}

0

1623382 LV 4 @ 2016-12-15 17:57:28

测试数据 #0: Accepted, time = 0 ms, mem = 496 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 496 KiB, score = 10
测试数据 #2: Accepted, time = 15 ms, mem = 496 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 496 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 496 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 496 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 496 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 492 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 500 KiB, score = 10
测试数据 #9: Accepted, time = 15 ms, mem = 496 KiB, score = 10
Accepted, time = 75 ms, mem = 500 KiB, score = 100
``C #include<stdio.h> int main() { int n,x,i,j,a,ans=0,mod=1; scanf("%d%d",&n,&x); for(i=1;i<=n;i++) { a=i; while(a>0) { if(a%10==x) ans++; a=a/10; } } printf("%d",ans); return 0; }

1623382 @ 2016-12-15 17:58:23

简单！！！！！！！！！！！！！

0

我是神 LV 10 @ 2016-11-18 10:03:45

#include<iostream>
#include<cstdio>
using namespace std;
int main(){
int n;
scanf("%d",&n);
int x;
scanf("%d",&x);
int count=0;
for(int i=1;i<=n;i++){
int j=i;
while(j!=0){
if(j%10==x){
count++;
}
j/=10;
}
}
cout<<count<<endl;
return 0;
}

0

未命名刺客 LV 8 @ 2016-11-14 20:29:25

宇宙第一水题
var i,j,n,x,t:longint;
begin
readln(n,x);t:=0;
for i:=1 to n do begin
j:=i;
while j>0 do begin
if j mod 10=x then t:=t+1;
j:=j div 10;
end;
end;
writeln(t);
end.

0

Rhine LV 7 @ 2016-10-20 14:47:16

#include<cstdio>
#include<cstring>
using namespace std;
int main(){
    int n,x,wei[9],num(0);
    scanf("%d%d",&n,&x);
    for(int i(1);i<=n;i++){
        int k(i),j=1;
        while(k){
            wei[j]=k%10;
            (x==wei[j++])?num++:0;
            k/=10;
        }
        memset(wei,0,sizeof(wei));***记得清零***
    }
    printf("%d\n",num);
    return 0;
}

Rhine @ 2016-10-20 14:47:52

记得清零

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