Codefroces #404 Div2

分析:把多面体和面数一一对应即可

 1 #include<iostream>
 2 #include<map>
 3 #include<cstring>
 4 #include<cstdio>
 5 using namespace std;
 6 int main()
 7 {
 8     int T;
 9     while(cin>>T){
10         map<string,int>mp;
11         mp["Tetrahedron"]=4;
12         mp["Cube"]=6;
13         mp["Octahedron"]=8;
14         mp["Dodecahedron"]=12;
15         mp["Icosahedron"]=20;
16         long long sum=0;
17         for(int i=0;i<T;i++){
18             string s;
19             cin>>s;
20             sum+=mp[s];
21         }
22         cout<<sum<<endl;
23     }
24     return 0;
25 }

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B题

分析:分别统计最早的r1和最晚的l2,以及最早的r2和最晚的l1，求二者的最大值即可

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 using namespace std;
 6 const int maxn=200000+10;
 7 const int INF=100000000;
 8 typedef struct
 9 {
10     long long l,r;
11 }point;
12 point p1[maxn],p2[maxn];
13 bool cmp1(point a,point b){
14     return a.r<b.r;
15 }
16 bool cmp2(point a,point b){
17     return a.l>b.l;
18 }
19 int n,m;
20 int main()
21 {
22     while(cin>>n){
23         long long minx=INF;
24         for(int i=0;i<n;i++){
25             scanf("%I64d%I64d",&p1[i].l,&p1[i].r);
26         }
27         sort(p1,p1+n,cmp1);
28         cin>>m;
29         for(int i=0;i<m;i++){
30             scanf("%I64d%I64d",&p2[i].l,&p2[i].r);
31         }
32         sort(p2,p2+m,cmp2);
33         int i=0,j=0;
34         long long ans=0;
35         long long cnt=p2[j].l-p1[i].r;
36         ans=max(ans,cnt);
37         sort(p1,p1+n,cmp2);
38         sort(p2,p2+m,cmp1);
39         long long h=0;
40         h=max(h,(p1[j].l-p2[i].r));
41         ans=max(h,ans);
42         cout<<ans<<endl;
43     }
44     return 0;
45 }

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C题

分析:若对于m>n，直接就是n天，若n>m，则前m天是必须的，后面相当于去寻找一个最小的x，使1+2+3+......+x>(n-m)，所以二分答案即可

 1 #include "iostream"
 2 #include "cstdio"
 3 #include "cstring"
 4 using namespace std;
 5 long long n,m;
 6 int main()
 7 {
 8     while(cin>>n>>m){
 9         if(m>n){
10             cout<<n<<endl;
11             continue;
12         }
13         n-=m;
14         long long l=0,r=2e9;
15         while(l<r){
16             long long mid=(l+r)>>1;
17             long long ans=(1+mid)*mid/2;
18             if(ans>=n) r=mid;
19             else l=mid+1;
20         }
21         cout<<m+l<<endl;
22     }
23     return 0;
24 }

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