CF1088C Ehab and a 2-operation task 構造
阿新 • • 發佈:2019-01-20
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最簡單的就是變成1~n的序列;
我們可以先加一個極大的數防止出現負數;
然後對於每一個%(a[ i ] - i);
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<functional> #include<sstream> //#include<cctype> //#pragma GCC optimize(2) using namespace std; #define maxn 1000005 #define inf 0x7fffffff //#define INF 1e18 #define rdint(x) scanf("%d",&x) #define rdllt(x) scanf("%lld",&x) #define rdult(x) scanf("%lu",&x) #define rdlf(x) scanf("%lf",&x) #define rdstr(x) scanf("%s",x) typedef long long ll; typedef unsigned long long ull; typedef unsigned int U; #define ms(x) memset((x),0,sizeof(x)) const long long int mod = 1e9 + 7; #define Mod 1000000000 #define sq(x) (x)*(x) #define eps 1e-4 typedef pair<int, int> pii; #define pi acos(-1.0) //const int N = 1005; #define REP(i,n) for(int i=0;i<(n);i++) typedef pair<int, int> pii; inline ll rd() { ll x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == ‘-‘) f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x; } ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); } int sqr(int x) { return x * x; } /*ll ans; ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans; } */ int n; int a[maxn]; int main() { //ios::sync_with_stdio(0); cin >> n; for (int i = 1; i <= n; i++)rdint(a[i]); cout << n + 1 << endl; cout << 1 << ‘ ‘ << n << ‘ ‘ << 99999 << endl; for (int i = 1; i <= n; i++) { a[i] += 99999; cout << 2 << ‘ ‘ << i << ‘ ‘ << a[i] - i << endl; a[i] %= (a[i] - i); // cout << a[i] << endl; } return 0; }
CF1088C Ehab and a 2-operation task 構造