1049. Counting Ones (30) --看
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代码容易超时,解决方法:
先用100作为最小单位,0~99内有20个1
100~199有20+100*(100中1个数)
200~299有20+100*(200中1个数)依次类推,发现还超时
将单位扩大到1000同上计算发现还超时,扩大到10000时不超时
#define _CRT_SECURE_NO_WARNINGS
#include<string>
#include<string.h>
#include<vector>
#include<map>
#include<stack>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<stdio.h>
using namespace std;
int count(int num )
{
int coun = 0;
while (num > 0)
{
if (num % 10 == 1)
{
coun++;
}
num /= 10;
}
return coun;
}
int main()
{
int N;
cin >> N;
int iCount = 0;
int iC = count(N / 10000);
iCount += iC;
if (0 != N % 10000)
{
for (int i = 1; i <= N % 10000; i++)
{
iCount = iCount + iC + count(i);
}
}
for (int i = N / 10000 * 10000; i >0;)
{
int iC = count(i - 10000);
iCount += (4000+10000*iC);
i -= 10000;
}
cout << iCount;
return 0;
}
另外一种方法:按照每位数来计算
#define _CRT_SECURE_NO_WARNINGS
#include<string>
#include<string.h>
#include<vector>
#include<map>
#include<stack>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<stdio.h>
using namespace std;
int main()
{
int N,M;
cin >> N;
M = N;
int sum = 0;
int i = 1;
while (N>0)
{
if (N%10 == 0)
sum=sum+ i*(N / 10);
else if (N%10 > 1)
sum =sum+ i*(N / 10 + 1);
else
sum =sum+ (i*(N / 10) + M%i+1);
i *= 10;
N /= 10;
}
cout << sum;
return 0;
}
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