2021CCPC湘潭全国邀请赛题解

Posted 2022-01-31 CleverXun

题目pdf下载
提取码: abcd
一共A了5题，j题签到题没出来，如果罚时少一点并且把J也写了的话，也许就金了

A. A+B Problem

题意：

给定a和b，求a+b,a+b如果大于1023或者小于-1024，就自动溢出

思路：模拟

比如1023+1=1024，1024>1023 就自动溢出成-1024，模拟即可

时间复杂度：O n

#include<bits/stdc++.h>
#define fer(i,a,b) for(re i = a ; i <= b ; ++ i)
#define re register int
#define pll pair<int,int> 
#define x first 
#define y second 
#define sf(x) scanf("%d",&x)
#define sfl(x) scanf("%lld",&x)
typedef long long ll ;
using namespace std;
const int N =  1e6 + 10 , M = 1010 , inf = 0x3f3f3f3f , mod = 1e9 + 7 ;

int main()

    int t ;
    cin >> t ;
    while(t--)
    
        int a , b ;
        cin >> a >> b ;
        int k = a + b ;
        if(k > 1023) k -= 2048 ;
        else if(k < -1024) k += 2048 ;
        cout << k << "\\n" ;
    
    return 0;

C. Calculate

题意：

给定x1,x2,y1,y2
求
答案对1e9+7取模

思路：数论分块

一维数论整除分块
第4个式子的答案有点问题
对其每一个区间左端点为l，右端点为r，值为cnt
答案为(r-l+1) $\times$ cnt $\times$ ${\sum }_{i=x1}^{x2}\lfloor \frac{i}{x1}\rfloor$
在考虑一下 求
${\sum }_{i=x1}^{x2}\lfloor \frac{i}{x1}\rfloor$

可以发现，
当x1<=i<2*x1 ，i/x1等于1
当2*x1<=i<3*x1,i/x1等于2
...........
当n*x1<=i<(n+1)*x1 , i/x1等于n
因此答案为 x1 * (1 + 2 + ........ + n ) + 大于等于(n+1)*x1的部分暴力求
在考虑一下1式
答案为 x1 * (1^2 + 2^2 + ........ + n^2) + 大于等于(n+1)*x1的部分暴力求
中间括号的部分用公式o1求即可

时间复杂度：O t$\sqrt{n}$

#include <iostream>
using namespace std;
typedef long long ll;
const int  mod = 1e9 + 7;

ll qpow(ll a, ll b) 
	ll res = 1;
	while (b) 
		if (b & 1)
			res = res * a % mod;
		a = a * a % mod;
		b >>= 1;
	
	return res;

ll inv6 = qpow(6, mod - 2);
ll inv2 = qpow(2, mod - 2);

ll cal_a2(ll x1, ll x2) 
	ll a = 0;
	ll k = x2 / x1 - 1;
	ll k2 = x2 / x1;
	ll k1 = x2 % x1 + 1;
	a = x1 * k % mod * (k + 1) % mod * (2 * k + 1) % mod * inv6 % mod;
	a = (a + k1 % mod * k2 % mod * k2 % mod) % mod;
	return a;


ll cal_b2(ll x1, ll x2) 
	ll b = 0;
	for (ll l = x1, r; l <= x2; l = r + 1) 
		r = x2 / (x2 / l);
		b += (r - l + 1) * (x2 / l) % mod * (x2 / l) % mod;
		b %= mod;
	
	return b;


ll cal_a(ll x1, ll x2) 
	ll a1 = 0;
	ll k = x2 / x1 - 1;
	ll k2 = x2 / x1;
	ll k1 = x2 % x1 + 1;
	a1 = (a1 + x1 * k % mod * (1 + k) % mod * inv2 % mod) % mod;
	a1 = (a1 + k1 * k2 % mod) % mod;
	return a1;


ll cal_b(ll x1, ll x2) 
	ll b = 0;
	for (ll l = x1, r; l <= x2; l = r + 1) 
		r = x2 / (x2 / l);
		b += (r - l + 1) * (x2 / l);
		b %= mod;
	
	return b;


ll cal_ab(ll x1, ll x2) 
	ll ab = 0;
	for (ll l = x1, r; l <= x2; l = r + 1) 
		r = x2 / (x2 / l);
		ll v = x2 / l;
		ll f = cal_a(x1, r) - cal_a(x1, l - 1);
		f = (f + mod) % mod;
		ab = (ab + v * f % mod) % mod;
	
	return ab;


int main() 
	int t;
	cin >> t;
	while (t--) 
		ll x1, x2, y1, y2;
		cin >> x1 >> x2 >> y1 >> y2;
		ll a1 = 0, b1 = 0, c1 = 0, d1 = 0;
		//b d que
		ll dx = x2 - x1 + 1;
		ll dy = y2 - y1 + 1;
		a1 = cal_a2(x1, x2) % mod;
		b1 = cal_b2(x1, x2) % mod;
		c1 = cal_a2(y1, y2) % mod;
		d1 = cal_b2(y1, y2) % mod;
		ll a = 0, b = 0, c = 0, d = 0;
		a = cal_a(x1, x2);
		c = cal_a(y1, y2);
		b = cal_b(x1, x2);
		d = cal_b(y1, y2);
		ll ab = 0, cd = 0;
		ab = cal_ab(x1, x2) * dy % mod;
		cd = cal_ab(y1, y2) * dx % mod;
		ll ans = 0;
		ans = (ans + (((dy * a1 % mod + dy * b1 % mod) % mod + dx * c1 % mod) % mod + dx * d1 % mod) % mod) % mod;
		ans = (ans + 2 * ab % mod) % mod;
		ans = (ans + 2 * a * c % mod) % mod;
		ans = (ans + 2 * a * d % mod) % mod;
		ans = (ans + 2 * b * c % mod) % mod;
		ans = (ans + 2 * b * d % mod) % mod;
		ans = (ans + 2 * cd % mod) % mod;
		ans %= mod;
		printf("%lld\\n", ans);
	
	return 0;

E. CCPC Strings

题意：

给一个整数n，会有2^n个不同的cp字符串
求每个不同字符串的ccpc子串的个数的总和
答案对1e9+7取膜

思路：bm算法求线性通项公式

先暴力预处理前20项
然后在用bm算法求出线性递推方程
o1输出即可

时间复杂度：20 * 2^20 + t

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1)if(b&1)res=res*a%mod;a=a*a%mod;return res;
// head
 
int _;
ll n;
namespace linear_seq 
    const int N=10010;
    ll res[N],base[N],_c[N],_md[N];
 
    vector<int> Md;
    ll mul(ll *a,ll *b,int k) 
        rep(i,0,k+k) _c[i]=0;
        rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (int i=k+k-1;i>=k;i--) if (_c[i])
            rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        rep(i,0,k) a[i]=_c[i];
    
    ll solve(ll n,VI a,VI b)  // a 绯绘暟 b 鍒濆€?b[n+1]=a[0]*b[n]+...
//        printf("%d\\n",SZ(b));
        ll ans=0,pnt=0;
        int k=SZ(a);
        assert(SZ(a)==SZ(b));
        rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
        Md.clear();
        rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
        rep(i,0,k) res[i]=base[i]=0;
        res[0]=1;
        while ((1ll<<pnt)<=n) pnt++;
        for (int p=pnt;p>=0;p--) 
            mul(res,res,k);
            if ((n>>p)&1) 
                for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
                rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            
        
        rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
        if (ans<0) ans+=mod;
        return ans;
    
    VI BM(VI s) 
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