Educational Codeforces Round 85 (Rated for Div. 2)
题目链接:http://codeforces.com/contest/1334/
A
思路:按照题意模拟下就行了,注意情况不要漏考虑,差值的比较
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1 //------------------------------------------------- 2 //Created by HanJinyu 3 //Created Time :二 4/21 21:10:17 2020 4 //File Name :85A.cpp 5 //------------------------------------------------- 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <list> 16 #include <map> 17 #include <string> 18 #include <math.h> 19 #include <stdlib.h> 20 #include <time.h> 21 using namespace std; 22 typedef double db; 23 typedef long long ll; 24 const int maxn = 200005; 25 26 int main() 27 { 28 //freopen("in.txt","r",stdin); 29 //freopen("out.txt","w",stdout); 30 int t; 31 scanf("%d",&t); 32 while(t--) 33 { 34 int n; 35 scanf("%d",&n); 36 int a[maxn],b[maxn]; 37 bool flag=false; 38 for(int i=0;i<n;i++){ 39 scanf("%d%d",&a[i],&b[i]); 40 if(b[i]>a[i]) 41 flag=true; 42 if(i!=0&&b[i]<b[i-1]) 43 flag=true; 44 if(i!=0&&a[i]<a[i-1]) 45 flag=true; 46 if(i!=0&&a[i]>a[i-1]&&b[i]<b[i-1]) 47 flag=true; 48 if(i!=0&&b[i]>b[i-1]&&a[i]<=a[i-1]) 49 flag=true; 50 if(i!=0&&((b[i]-b[i-1])>(a[i]-a[i-1]))) 51 flag=true; 52 } 53 if(flag) 54 printf("NO\n"); 55 else 56 printf("YES\n"); 57 } 58 59 return 0; 60 }
B
思路:求和除以n看结果是不是大于等于x,如果是,那么直接输出n,其次,将数组降序排序,一位一位考虑,用和累加之前的值,当和除以当前值得位置+1即个数的时候,看这些个数的平均值是否大于等于x,如果小于x直接退出,当前位置+1即是所求个数了
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1 //------------------------------------------------- 2 //Created by HanJinyu 3 //Created Time :二 4/21 22:00:15 2020 4 //File Name :85B.cpp 5 //------------------------------------------------- 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <list> 16 #include <map> 17 #include <string> 18 #include <math.h> 19 #include <stdlib.h> 20 #include <time.h> 21 using namespace std; 22 typedef double db; 23 typedef long long ll; 24 const int maxn = 200005; 25 26 int main() 27 { 28 //freopen("in.txt","r",stdin); 29 //freopen("out.txt","w",stdout); 30 int t; 31 scanf("%d",&t); 32 while(t--){ 33 ll sum=0; 34 ll y[maxn]; 35 ll n,x; 36 scanf("%lld %lld",&n,&x); 37 for(int i=0;i<n;i++) 38 { 39 scanf("%lld",&y[i]); 40 sum+=y[i]; 41 } 42 if(sum/n>=x) 43 printf("%lld\n",n); 44 else{ 45 sort(y,y+n,greater<ll>()); 46 if(y[0]<x) 47 { 48 49 printf("0\n"); 50 continue; 51 } 52 ll ss=0; 53 ll r=0; 54 for(int i=0;i<n;i++) 55 { 56 ss+=y[i]; 57 if(ss/(i+1)>=x) 58 r=(i+1); 59 else 60 break; 61 } 62 printf("%lld\n",r); 63 } 64 } 65 66 return 0; 67 }
C
思路:脑子转不过来,用了一个比较好理解的方法,他是按照顺序来杀怪物的,那么用一个res数组来记录a[i]-b[i-1],当res[i]大于0的时候,将其累加起来,这个和算的是如果不杀作为第一个数所用子弹数的总和,然后枚举每种情况,杀a[i]时所要花费的子弹数,这时候就要减去自身的res[i]的值,因为跟上一个怪物的爆炸值没有关系,因此这个sum和里包含的res[i]值要减去,这时候是除了自身其他杀死怪物所要的子弹数,当然还要加上自身的a[i]值,这才是杀死当前怪物所用的所有子弹数,注意当res[i]<0的时候就是这个怪物可以完全被爆炸掉,减去res[i]的时候判断是否大于0就行,因为负的不在sum和之中,因此不用减去,最后所有情况求最小值即可。
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//------------------------------------------------- //Created by HanJinyu //Created Time :四 4/30 12:48:30 2020 //File Name :85C.cpp //------------------------------------------------- #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <list> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; typedef double db; typedef long long ll; const int maxn = 200005; int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int t,n; scanf("%d",&t); while(t--) { scanf("%d",&n); ll a[n],b[n]; for(int i=0;i<n;i++) scanf("%lld%lld",a+i,b+i); ll res[n],sum=0; for(int i=0;i<n;i++) { if(i==0) res[i]=a[i]-b[n-1]; else res[i]=a[i]-b[i-1]; if(res[i]>0) sum+=res[i]; } ll he=1e18; for(int i=0;i<n;i++) { he=min(he,sum+a[i]-max(0ll,res[i])); } printf("%lld\n",he); } return 0; }