北京信息科技大学第十一届程序设计竞赛

Problem A kotori和糖果

https://ac.nowcoder.com/acm/contest/940/A

题意：合并，求代价最小值

题解：将一个堆二分，递归求该堆合并的最小代价，用map判重。

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
map<ll,ll>mp;
ll sloved(ll x){
    if(mp[x])
        return mp[x];
    if(x==0||x==1||( x & (x - 1)) == 0 )
        return mp[x]=0;
    if(x&1)
        return mp[x]=sloved(x>>1)+sloved((x>>1)+1)+1;
    else
        return mp[x]=2*sloved(x>>1);
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    scanf("%lld",&t);
    while(t--){
        scanf("%lld",&n);
        cout<<sloved(n)<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

Problem B kotori和气球

https://ac.nowcoder.com/acm/contest/940/B

题意：求满足要求的n个气球的排列种数

题解：当仅当第一个可以选择n种颜色，其他不能与前面相同，所以只能选择m-1种颜色

C++版本一

题解：快速幂+规律+数学

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=109;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
ll powmod(ll a,ll b=MOD-2,ll c=MOD){
    ll res=1;
    ll base=a%c;
    while(b){
        if(b&1)res=(res*base)%c;
        base=(base*base)%c;
        b>>=1;
    }
    return res;
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%lld%lld",&n,&m);
    cout<<(n*powmod(n-1,m-1,MOD))%MOD<<endl;
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

题解：dp递推，在第i个位置放颜色为x的球的方案数为在i-1处放除x以外的球的方案数之和
状态转移方程：s[i][x]=sum{s[i-1][y]，y!=x}

北京信息科技大学第十一届程序设计竞赛

Problem A kotori和糖果

Problem B kotori和气球

Problem C kotori和出道

Problem D kotori和迷宫

Problem E kotori和素因子

Problem F kotori和n皇后

Problem G kotori和抽卡（二）

Problem H andy和购物

Problem I andy种树

Problem J andy的树被砍了

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