找规律的题目主要分为两大类
找规律的题目主要分为两大类
没有参照的找规律
2018南京ICPC G 题
写出暴力程序，发现相邻项之间做除法好像有某种规律，给所有的ans都乘上24发现是相邻的四项的乘积
#include <bits/stdc++.h>
using namespace std;
struct x{
double x,y;
}d[100000];
int cnt = 0;
double esp = 1e-4;
double Slove(int a,int b){
return (d[a].x - d[b].x) * (d[a].x - d[b].x) + (d[a].y - d[b].y) * (d[a].y - d[b].y);
}
double g3 = sqrt(3.0) * 1.0;
inline int s(int n){
int c = 0;
cnt = 0;
for(int i = n + 1;i;i --){
x t = (x){
c * 1.0,c * 1.0 * g3};
for(int j = 0;j < i;j ++){
d[++cnt] = {
j * 2.0,0};
d[cnt].x += t.x;
d[cnt].y += t.y;
}
c ++;
}
//for(int i = 1;i <= cnt;i ++) printf("[%f %f] ",d[i].x,d[i].y);
int ans = 0;
for(int i = 1;i <= cnt;i ++){
for(int j = i + 1;j <= cnt;j ++){
for(int k = j + 1;k <= cnt;k ++){
if(abs(Slove(i,j) - Slove(j,k)) <= esp && abs(Slove(i,k) - Slove(j,k)) <= esp ){
ans ++;
//printf("[%d %d %d]\n",i,j,k);
}
}
}
}
//printf("%d\n",ans);
return ans;
}
inline biao(){
int a[100];
int c = 0;
printf(" n-5");
for(int i = -4;i <= 5;i ++){
if(i <0)
printf(" n%d",i);
else printf(" n+%d",i);
}
for(int i = -5;i <= 5;i ++){
if(i < 0)
printf(" 2n%d",i);
else printf(" 2n+%d",i);
}
puts("");
for(int i = 1;i <= 10;i ++){
//printf("%d ",i);
a[i] = s(i);
a[i] *= 24;
for(int j = i - 5;j <= i + 5;j ++){
if(j <= 0) printf(" |e");
else if(a[i] % j == 0) printf(" |o");
else printf(" |X");
}
for(int j = 2 * i - 5;j <= 2 * i + 5;j ++){
if(j <= 0) printf(" |e");
else if(a[i] % j == 0) printf(" |o");
else printf(" |X");
}
puts("");
}
}
#define ll long long
const ll Mod = 1e9 + 7;
ll Qpow(ll a,ll b){
ll ans = 1,base = a;
while(b){
if(b & 1){
ans *= base;
ans %= Mod;
}
base *= base;
base %= Mod;
b >>= 1;
}
return ans;
}
int main(){
int _;scanf("%d",&_);
ll inv = Qpow(24,Mod - 2);
while(_--){
ll n;scanf("%lld",&n);
ll ans = n * (n + 1) % Mod;
ans *= (n + 2);
ans %= Mod;
ans *= (n + 3);
ans %= Mod;
ans *= inv;
ans %= Mod;
printf("%lld\n",ans);
}
return 0;
}