Codeforces Educational Rounds 85 A~C

A:Level Statistics

题意:统计n个游戏数据,p代表游玩次数,c代表通关次数,每次游玩都不一定通关,求这些数据是否合法

题解:1.游玩次数不能小于通关次数   2.游玩次数和通关次数必须单增  3.每次增加的游玩次数不能小于通关次数

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <unordered_set>
#include <unordered_map>
#define ll long long
#define fi first
#define se second
#define pb push_back
#define me memset
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
using namespace std;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;

int t;
int n;
int p,c,mp=0,mc=0,tp=0,tc=0;

int main() {
    ios::sync_with_stdio(false);
    cin>>t;
    while(t--){
        tp=0,tc=0;
        bool flag=1;
        cin>>n;
        for(int i=0;i<n;++i){
            cin>>p>>c;
            if(i==0) tp=p,tc=c;
            if(p<tp || c<tc || (p==tp && c>tc) || p<c || (p-tp<c-tc)){
                flag=0;
            }
            tp=p;
            tc=c;
        }
        if(flag==0) printf("NO\n");
        else printf("YES\n");
    }

    return 0;
}

 

B. Middle Class

题意:定义每人拥有的钱不小于x时为富人,现在有n个人,可以选(1<=x<=n)个人出来将他们的财产平分给所有人,求最多能有多少富人

题解:对每个人拥有的钱排个序,然后贪心,从最有钱的人开始往前遍历,记录一个最大值即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <unordered_set>
#include <unordered_map>
#define ll long long
#define fi first
#define se second
#define pb push_back
#define me memset
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
using namespace std;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
 
int t;
int n;
long double x,sum[N];
int ans=0;
long double a[N];
int main() {
    ios::sync_with_stdio(false);
    cin>>t;
     while(t--){
        cin>>n>>x;
        ans=0;
         for(int i=0;i<n;++i){
             cin>>a[i];
         }
        sort(a,a+n);
         for(int i=n-1;i>=0;--i){
             sum[n-i]=sum[n-i-1]+a[i];
         }
        for(int i=1;i<=n;++i){
            long double tmp=sum[i]/i;
            if(tmp>=x){
                ans=max(ans,i);
            }
        }
        printf("%d\n",ans);
     }
 
 
 
    return 0;
}

 

C. Circle of Monsters

题意:有n个怪物围成一个圈,每次攻击可以对怪物造成一点伤害,当第i个怪物扑街后会对后面一个(n的后面是1)怪物造成b[i]点伤害(如果i+1个怪物存活的话),求最少攻击多少次能将怪物全部消灭

题解:枚举每个怪物受到前一个怪物的阵亡伤害,然后按顺序遍历一边求个最小值就行了(建议用scanf和pritnf,容易T)

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <unordered_set>
#include <unordered_map>
#define ll long long
#define fi first
#define se second
#define pb push_back
#define me memset
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
using namespace std;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
 
int t;
ll n,a[N],b[N],c[N];
ll sum;
ll ans=1e18+10;
int main() {
    scanf("%d",&t);
    while(t--){
        scanf("%lld",&n);
        ans=1e18+10;
        sum=0;
        for(int i=1;i<=n;++i){
            scanf("%lld %lld",&a[i],&b[i]);
            if(i>1) c[i]=max((ll)0,a[i]-b[i-1]);
        }
        c[1]=max((ll)0,a[1]-b[n]);
        if(n==1){
            printf("%lld\n",a[1]);
            continue;
        }
        for(int i=1;i<=n;++i){
           sum+=c[i];
        }
        for(int i=1;i<=n;++i){
            ans=min(ans,a[i]+sum-c[i]);
        }
        printf("%lld\n",ans);
    }
 
    return 0;
}