HDU 1337 && POJ 1218&& zju 1350 Method Summary

Topic Link http://acm.hdu.edu.cn/showproblem.php?pid=1337 Hangzhou Electric

http://poj.org/problem?id=1218 Tsinghua

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1350 Zhejiang

Originally thought that a small water problem is not worth summarizing, but found that there are many ways, good magic AH

Method One: The data is very small simulation is good

#include <iostream> #include <algorithm> #include <stdio.h> #include <string.h>using namespace Std;int A[110];int Main () {    int t, N;    scanf ("%d", &t);    while (t--)    {        memset (a, 0, sizeof (a));        scanf ("%d", &n);        for (int i=1, i<=n; i++)        {for            (int j=i; j<=n; j+=i)            {                a[j]=a[j]^1;            }        }        int ans=0;        for (int i=1; i<=n; i++)        {            if (a[i]==1)                ans++;        }        printf ("%d\n", ans);    }    return 0;}

Method Two: In fact, the prisoner can escape is a regular follow, the door has two forms, as long as the total number of prisoners to open and close is odd, such as 3 1 open 3 off, the sum of the switch is even; for example, 4 1 open 2 off 4 open, the sum of the switch is odd is to see whether there are odd factors;

#include <iostream>#include<algorithm>#include<stdio.h>#include<string.h>using namespacestd;intSolveintx) {    intCnt=0;  for(intI=1; i<=x; i++)    {        if(x%i==0) CNT++; }    if(cnt%2==0)        return 0; Else        return 1;}intMain () {intT, N; scanf ("%d", &t);  while(t--) {scanf ("%d", &N); intans=0;  for(intI=1; i<=n; i++)        {           if(Solve (i)) ans++; } printf ("%d\n", ans); }    return 0;}

The Third kind: is to see others to know, really is the world of great wonders;

The principle is: only the complete square number can have an odd number of factors, such as 3, 9, so only need to ask for the total number of squares within n;

The number of complete squares within n =sqrt (n);

#include <cstdio> #include <cmath>int main () {    int t,n;    scanf ("%d", &t);    while (t--)    {        scanf ("%d", &n);        printf ("%d\n", (int) sqrt (n));    }    return 0;}

　　

HDU 1337 && POJ 1218&& zju 1350 Method Summary