【noip2014】day2

T1

[前缀和,差分]

求二维前缀和然后大概差分一下就好了?

 【code】

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
#define File "wireless"
inline void file(){
    freopen(File".in","r",stdin);
    freopen(File".out","w",stdout);
} 
inline int read(){
    int x = 0,f = 1; char ch = getchar();
    while(ch < '0' || ch > '9'){if(ch=='-')f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9'){x = (x<<1) + (x<<3) + ch-'0'; ch = getchar();}
    return x*f;
}
const int mxn = 130;
int d,n;
int s[mxn][mxn];
int a[mxn][mxn];

inline int Q(int x_1,int y_1,int x_2,int y_2){
    return s[x_2][y_2] - s[x_1-1][y_2] - s[x_2][y_1-1] + s[x_1-1][y_1-1];
}
int cnt,ans;
int main(){
    file();
    d = read(),n = read();
    for(int i = 1;i <= n; ++i){
        int x,y,k;
        x = read(),y = read(),k = read();
        a[x][y] = k;
    }
    
//    s[0][0] = a[0][0];
    for(int i = 0;i <= 128; ++i)
        for(int j = 0;j <= 128; ++j)
            s[i][j] = (s[i][j-1] + s[i-1][j] - s[i-1][j-1] + a[i][j]);
//            ,printf("%d\n",s[i][j]);


//    printf("%d\n",ans);
    for(int i = 0;i <= 128; ++i){
        for(int j = 0;j <= 128; ++j){
            int x_1,y_1,x_2,y_2;
            x_1 = (i-d >= 0) ? (i-d) : 0;
            y_1 = (j-d >= 0) ? (j-d) : 0;
            x_2 = (i+d <= 128) ? (i+d) : 128;
            y_2 = (j+d <= 128) ? (j+d) : 128;
            int t = Q(x_1,y_1,x_2,y_2);
            if(t == ans) cnt++;
            else if(t > ans) ans = t,cnt = 1;
        }
    }
    printf("%d %d\n",cnt,ans);        
    return 0;
}
View Code

 

T2

[Bfs]

通过建反图判断每个点是否能到达终点。

通过枚举每个能达到终点的点,判断与之相连的边连接的点是否能出现在路径中。

最后bfs求出起点到终点的距离即可。                

【code】

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
#define File "road"
inline void file(){
    freopen(File".in","r",stdin);
    freopen(File".out","w",stdout);
} 
inline int read(){
    int x = 0,f = 1; char ch = getchar();
    while(ch < '0' || ch > '9'){if(ch=='-')f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9'){x = (x<<1) + (x<<3) + ch-'0'; ch = getchar();}
    return x*f;
}
const int mxn = 1e4 + 10;
int n,m;
bool v1[mxn],v2[mxn];//是否能出现在路途中。是否能到达终点。 
int dis[mxn];
vector<int>E1[mxn];
vector<int>E2[mxn];
#define pb push_back
int st,ed;
queue<int> q;

int main(){
//    file();
    n = read(),m = read();
    for(int i = 1;i <= m; ++i){
        int x = read(),y = read();
        E1[x].pb(y),E2[y].pb(x);
    }    
    st = read(),ed = read();
    
    v2[ed] = 1; q.push(ed);
    while(q.size()){
        int x = q.front();
        q.pop();
        for(int i = E2[x].size()-1; i >= 0; --i){
            int y = E2[x][i];
            if(!v2[y]){
                q.push(y);
                v2[y] = 1;
            }
        }
    }
    if(!v2[st]){
        puts("-1");
        return 0;
    }
    for(int i = 1;i <= n; ++i){
        if(v2[i]){
            v1[i] = 1;
            for(int j = E1[i].size()-1; j >= 0; --j){
                int y = E1[i][j];
                if(!v2[y]){
                    v1[i] = 0;
                    break;
                }                
            }
        }
    }
    
    dis[st] = 1; q.push(st);
    while(q.size()){
        int x = q.front();
        q.pop();
        if(x == ed){
            printf("%d\n",dis[ed]-1);
            return 0;
        }
        for(int i = E1[x].size()-1; i >= 0; --i){
            int y = E1[x][i];
            if(v1[y] && !dis[y]){
                dis[y] = dis[x] + 1;
                q.push(y);
            }
        }
    }
    
    puts("-1");
    return 0;
}
View Code

 

T3

[数学]

读入取模。秦九韶公式。

【code】 

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
#define File "equation"
inline void file(){
    freopen(File".in","r",stdin);
    freopen(File".out","w",stdout);
}
const int mod = 998244353;
const int mxn = 110;
inline ll read(){
    ll x = 0,f = 1; char ch = getchar();
    while(ch < '0' || ch > '9'){if(ch=='-')f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9'){x = ((x<<1)%mod + (x<<3)%mod + ch-'0')%mod; ch = getchar();}
    return x*f; 
}
ll n,m;
ll a[mxn],b[mxn];
ll tot(0);
inline bool J(ll x){
    ll s(0);
    for(ll i = n; i; --i)
        s = ((s + a[i])*x) % mod;
    s = (s+a[0]) % mod;
    return s==0;
}//秦九韶公式的判断。返回得到的s是否等于0 

int main(){
    file();
    n = read(),m = read();    
    for(ll i = 0;i <= n; ++i) a[i] = read();
    for(ll i = 1;i <= m; ++i){
        if(J(i)) 
            b[++tot] = i;//记录答案 
    }
    printf("%lld\n",tot);
    for(ll i = 1;i <= tot; ++i) printf("%lld\n",b[i]);
    return 0;
}
View Code

 

我又预感因为交题交得慢我今天要ak掉rating。

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转载自www.cnblogs.com/ve-2021/p/11364388.html
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