The 5th Zhejiang Provincial Collegiate Programming Contest(ZOJ2965

本文链接：https://blog.youkuaiyun.com/piaocoder/article/details/51220982

本文精选了多个ACM编程比赛题目并提供了详细的解题思路与AC代码，涉及字符串处理、模拟、组合数学等算法，适合算法初学者及竞赛选手参考。

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Accurately Say "CocaCola"!

题目链接：

http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=2965

解题思路：

暴力打表即可。。。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int vis[805];
int cnt[805];

int main(){
    memset(vis,0,sizeof(vis));
    memset(cnt,0,sizeof(cnt));
    for(int i = 1; i*7 <= 800; ++i)
        vis[i*7] = 1;
    for(int i = 1; i <= 800; ++i){
        int x = i;
        while(x){
            if(x%10 == 7){
                vis[i] = 1;
                break;
            }
            x /= 10;
        }
    }
    int flag = 0;
    for(int i = 800; i >= 1; --i){
        if(vis[i] && flag)
            cnt[i] = cnt[i+1]+1;
        else if(vis[i]){
            cnt[i] = 1;
            flag = 1;
        }
        else
            flag = 0;

    }
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);
        for(int i = 1; i <= 800; ++i){
            if(cnt[i] >= n){
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}

Build The Electric System

题目链接：

http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=2966

解题思路：

最小生成树。。。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 1000005;
int n,m;
int pa[505];

struct Edge{
    int u,v,w;
}edge[N];


bool cmp(Edge a,Edge b){
    return a.w < b.w;
}

int findset(int x){
    if(pa[x] != x)
        pa[x] = findset(pa[x]);
    return pa[x];
}

int kruskal(){
    int cnt = n,sum = 0;
    for(int i = 0; i < m; i++){
        int u = findset(edge[i].u);
        int v = findset(edge[i].v);
        if(u != v){
            sum += edge[i].w;
            pa[v] = u;
            if(--cnt == 0)
                break;
        }
    }
    return sum;
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        for(int i = 0; i <= n; ++i)
            pa[i] = i;
        int u,v,w;
        for(int i = 0; i < m; ++i)
            scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
        sort(edge,edge+m,cmp);
        printf("%d\n",kruskal());
    }
    return 0;
}

Easy Task

题目链接：

http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=2969

解题思路：

简单模拟即可。。。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int N = 1005;
int n,m;
int arr[N];
int arr2[N];

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        int tt = n+1; int sum = 0;
        for(int i = 0; i < tt; ++i){
            scanf("%d",&arr[i]);
            if(i != n)
                sum += arr[i];
        }
        if(sum == 0)
            printf("0\n");
        else{
            for(int i = 0; i < tt; ++i){
                int l = tt - i - 1;
                arr2[i] = arr[i] * l;
            }
            for(int i = 0; i < n-1; ++i)
                printf("%d ",arr2[i]);
            printf("%d\n",arr2[n-1]);
        }
    }
    return 0;
}

Faster, Higher, Stronger

题目链接：

http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=2970

解题思路：

水题。。。

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

char str[1005];
int arr[100005];

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%s",str);
        int n;
        scanf("%d",&n);
        for(int i = 0; i < n; ++i)
            scanf("%d",&arr[i]);
        sort(arr,arr+n);
        if(str[0] == 'F')
            printf("%d\n",arr[0]);
        else
            printf("%d\n",arr[n-1]);
    }
    return 0;
}

Give Me the Number

题目链接：

http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=2971

解题思路：

模拟即可。。。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;

map<string,int> ma;

int main(){
    ma.clear();
    ma["zero"] = 0; ma["one"] = 1; ma["two"] = 2; ma["three"] = 3; ma["four"] = 4;
    ma["five"] = 5; ma["six"] = 6; ma["seven"] = 7; ma["eight"] = 8; ma["nine"] = 9;
    ma["ten"] = 10; ma["eleven"] = 11; ma["twelve"] = 12; ma["thirteen"] = 13; ma["fourteen"] = 14;
    ma["fifteen"] = 15; ma["sixteen"] = 16; ma["seventeen"] = 17; ma["eighteen"] = 18; ma["nineteen"] = 19;
    ma["twenty"] = 20; ma["thirty"] = 30; ma["forty"] = 40; ma["fifty"] = 50; ma["sixty"] = 60;
    ma["seventy"] = 70; ma["eighty"] = 80; ma["ninety"] = 90;
    int T;
    scanf("%d",&T);
    getchar();
    while(T--){
        string str,tmp = "";
        getline(cin,str);
        int len = str.size();
        int ans = 0,tmpans = 0;
        for(int i = 0; i <= len; ++i){
            if(i < len && str[i] != ' ')
                tmp += str[i];
            else{
                //cout<<tmp<<endl;
                if(tmp == "million")
                    ans += tmpans*1000000, tmpans = 0;
                else if(tmp == "thousand")
                    ans += tmpans*1000, tmpans = 0;
                else if(tmp == "hundred")
                     tmpans *= 100;
                else if(ma[tmp])
                    tmpans += ma[tmp];
                tmp = "";
                //cout<<tmpans<<endl;
            }
        }
        ans += tmpans;
        printf("%d\n",ans);
    }
    return 0;
}

Kinds of Fuwas

题目链接：

http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=2975

解题思路：

组合数学。。。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

char maze[255][255];
int vis[255][255];
char fuwa[6] = "BJHYN";
int n,m;

int solve(char c){
    memset(vis,0,sizeof(vis));
    int cnt,res = 0;
    for(int i = 0; i < n-1; ++i){
        for(int j = 0; j < m-1; ++j){
            if(maze[i][j] == c){
                for(int k = j+1; k < m; ++k){
                    if(maze[i][k] == c && !vis[j][k]){
                        vis[j][k] = 1;
                        vis[k][j] = 1;
                        cnt = 1;
                        for(int p = i+1; p < n; ++p){
                            if(maze[p][j] == c && maze[p][k] == c)
                                ++cnt;
                        }
                        res += cnt*(cnt-1)/2;
                    }
                }
            }
        }
    }
    return res;
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        for(int i = 0; i < n; ++i)
            scanf("%s",maze[i]);
        int ans = 0;
        for(int i = 0; i < 5; ++i)
            ans += solve(fuwa[i]);
        printf("%d\n",ans);
    }
    return 0;
}

Light Bulbs

题目链接：

http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=2975

解题思路：

题目大意：

上面有几盏灯（灯可以看做是点），每盏灯都有一定的光照强度，求地上的能接受到最多光强的点，求这个最大的光强。

算法思想：

因为数据量不大，所以可以直接暴力枚举。

如果数据很大的话，可以使用模拟退火。

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

struct Point{
    double x,y,z;
    double I;
    Point(){}
    Point(double _x,double _y,double _z){x = _x; y = _y; z = _z;}
}po[105];
int vis[105][105];

double get_dis(Point p,Point q){
    return sqrt((p.x-q.x)*(p.x-q.x)+(p.y-q.y)*(p.y-q.y)+(p.z-q.z)*(p.z-q.z));
}

double Dis(int n,Point q){
    double dis = 0;
    for(int i = 0; i < n; i++){
        double R = get_dis(po[i],q);
        dis += po[i].I/(R*R)*(po[i].z/R);
    }
    return dis;
}

double solve(int n){
    double ans = 0;
    for(int i = -100; i <= 100; ++i){
        for(int j = -100; j <= 100; ++j){
            ans = max(ans,Dis(n,Point(i,j,0)));
        }
    }
    return ans;
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);
        for(int i = 0; i < n; i++)
            scanf("%lf%lf%lf%lf",&po[i].x,&po[i].y,&po[i].z,&po[i].I);
        printf("%.2lf\n",solve(n));
    }
    return 0;
}