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  • Heron and His Triangle HDU

    题目链接:https://vjudge.net/problem/HDU-6222

    思路:打表找规律。

     然后因为数据范围较大可以考虑用字符串模拟,或者__int128要注意用一个快读快输模板。

      1 #include <bits/stdc++.h>
      2 #include <time.h>
      3 #include <set>
      4 #include <map>
      5 #include <stack>
      6 #include <cmath>
      7 #include <queue>
      8 #include <cstdio>
      9 #include <string>
     10 #include <vector>
     11 #include <cstring>
     12 #include <utility>
     13 #include <cstring>
     14 #include <iostream>
     15 #include <algorithm>
     16 #include <list>
     17 using namespace std;
     18 //cout<<setprecision(10)<<fixed;
     19 #define eps 1e-6
     20 #define PI acos(-1.0)
     21 #define lowbit(x) ((x)&(-x))
     22 #define zero(x) (((x)>0?(x):-(x))<eps)
     23 #define mem(s,n) memset(s,n,sizeof s);
     24 #define rep(i,a,b) for(int i=a;i<=b;i++)
     25 #define rep2(i,a,b) for(int i=a;i>=b;i--)
     26 #define ios {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}
     27 typedef long long ll;
     28 typedef unsigned long long ull;
     29 const int maxn=1e6+5;
     30 const ll Inf=0x7f7f7f7f7f7f7f7f;
     31 const ll mod=1e9;
     32 //const int N=3e3+5;
     33 bool isPowerOfTwo(int n) { return n > 0 && (n & (n - 1)) == 0; }//判断一个数是不是 2 的正整数次幂
     34 int modPowerOfTwo(int x, int mod) { return x & (mod - 1); }//对 2 的非负整数次幂取模
     35 int getBit(int a, int b) { return (a >> b) & 1; }// 获取 a 的第 b 位,最低位编号为 0
     36 int Max(int a, int b) { return b & ((a - b) >> 31) | a & (~(a - b) >> 31); }// 如果 a>=b,(a-b)>>31 为 0,否则为 -1
     37 int Min(int a, int b) { return a & ((a - b) >> 31) | b & (~(a - b) >> 31); }
     38 ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
     39 ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
     40 inline int read()
     41 {
     42     int X=0; bool flag=1; char ch=getchar();
     43     while(ch<'0'||ch>'9') {if(ch=='-') flag=0; ch=getchar();}
     44     while(ch>='0'&&ch<='9') {X=(X<<1)+(X<<3)+ch-'0'; ch=getchar();}
     45     if(flag) return X;
     46     return ~(X-1);
     47 }
     48 inline void write(int X)
     49 {
     50     if(X<0) {X=~(X-1); putchar('-');}
     51     if(X>9) write(X/10);
     52     putchar(X%10+'0');
     53 }
     54 /*
     55 inline int write(int X)
     56 {
     57     if(X<0) {putchar('-'); X=~(X-1);}
     58     int s[20],top=0;
     59     while(X) {s[++top]=X%10; X/=10;}
     60     if(!top) s[++top]=0;
     61     while(top) putchar(s[top--]+'0');
     62 }
     63 */
     64 int Abs(int n) {
     65   return (n ^ (n >> 31)) - (n >> 31);
     66   /* n>>31 取得 n 的符号,若 n 为正数,n>>31 等于 0,若 n 为负数,n>>31 等于 -1
     67      若 n 为正数 n^0=n, 数不变,若 n 为负数有 n^(-1)
     68      需要计算 n 和 -1 的补码,然后进行异或运算,
     69      结果 n 变号并且为 n 的绝对值减 1,再减去 -1 就是绝对值 */
     70 }
     71 ll binpow(ll a, ll b) {
     72   ll res = 1;
     73   while (b > 0) {
     74     if (b & 1) res = res * a%mod;
     75     a = a * a%mod;
     76     b >>= 1;
     77   }
     78   return res%mod;
     79 }
     80 void extend_gcd(ll a,ll b,ll &x,ll &y)
     81 {
     82     if(b==0) {
     83         x=1,y=0;
     84         return;
     85     }
     86     extend_gcd(b,a%b,x,y);
     87     ll tmp=x;
     88     x=y;
     89     y=tmp-(a/b)*y;
     90 }
     91 ll mod_inverse(ll a,ll m)
     92 {
     93     ll x,y;
     94     extend_gcd(a,m,x,y);
     95     return (m+x%m)%m;
     96 }
     97 ll eulor(ll x)
     98 {
     99    ll cnt=x;
    100    ll ma=sqrt(x);
    101    for(int i=2;i<=ma;i++)
    102    {
    103     if(x%i==0) cnt=cnt/i*(i-1);
    104     while(x%i==0) x/=i;
    105    }
    106    if(x>1) cnt=cnt/x*(x-1);
    107    return cnt;
    108 }
    109 void scan(__int128 &x)//输入
    110 {
    111     x = 0;
    112     int f = 1;
    113     char ch;
    114     if((ch = getchar()) == '-') f = -f;
    115     else x = x*10 + ch-'0';
    116     while((ch = getchar()) >= '0' && ch <= '9')
    117         x = x*10 + ch-'0';
    118     x *= f;
    119 }
    120 void _print(__int128 x)
    121 {
    122     if(x > 9) _print(x/10);
    123     putchar(x%10 + '0');
    124 }
    125 __int128 a[maxn];
    126 int main()
    127 {
    128     int t=read();
    129     a[1]=4;
    130     a[2]=14;
    131     __int128 b=1e31;
    132     int i;
    133     for( i=3;;i++)
    134     {
    135         a[i]=a[i-1]*4-a[i-2];
    136         if(a[i]>b) break;
    137     }
    138     while(t--)
    139     {
    140         __int128 n;
    141         scan(n);
    142         for(int j=1;j<=i;j++)
    143         {
    144             if(a[j]>=n) {_print(a[j]);puts("");break;}
    145         }
    146     }
    147     return 0;
    148 }
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  • 原文地址:https://www.cnblogs.com/zpj61/p/13644802.html
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