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Codeforces Round #262 (Div. 2)

分类: IT文章 • 2022-07-29 00:33:10

A. Vasya and Socks http://codeforces.com/contest/460/problem/A

水题暴力

 1 #include<cstdio>
 2 int main(){
 3     int n,m;
 4     while(~scanf("%d%d",&n,&m)){
 5         int ans=0;
 6         while(n){
 7             n--;
 8             ans++;
 9             if(!(ans%m)) n++;
10         }
11         printf("%d
",ans);
12     }
13     return 0;
14 }

View Code

B. Little Dima and Equation http://codeforces.com/contest/460/problem/B

sx=x所有位的和，要知道sx在1~1e9的x 对应的是1~81，枚举81，判断。

 1 #include<cstdio>
 2 typedef __int64 LL;
 3 LL gxpow(int x,int n){
 4     LL res=1;
 5     while(n--) res*=x;
 6     return res;
 7 }
 8 int sum(LL x){
 9     int res=0;
10     while(x){
11         res+=x%10;
12         x/=10;
13     }
14     return res;
15 }
16 LL ans[128];
17 int main(){
18     int a,b,c;
19     while(~scanf("%d%d%d",&a,&b,&c)){
20         int la=0;
21         for(int i=1;i<=81;i++){
22             LL x=b*gxpow(i,a)+c;
23             if(0<x&&x<1e9&&sum(x)==i){
24                 ans[la++]=x;
25             }
26         }
27         printf("%d
",la);
28         if(!la) continue;
29         for(int i=0;i<la;i++){
30             printf("%d ",ans[i]);
31         }
32         puts("");
33     }
34     return 0;
35 }

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C. Present http://codeforces.com/contest/460/problem/C

二分答案，注意LR上下界，瞎定义tle了一次，对答案贪心的从前往后浇水，看能否构成答案，区间更新on左++右--是线段树nlogn的10分之一的时间46ms vs 390ms。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define mt(a,b) memset(a,b,sizeof(a))
 5 using namespace std;
 6 const int M=100010;
 7 int n,m,w,a[M],lazy[M];
 8 bool judge(int ans){
 9     mt(lazy,0);
10     int add=0,left=m;
11     for(int i=1;i<=n;i++){
12         add+=lazy[i];
13         int now=a[i]+add;
14         if(now<ans){
15             int need=ans-now;
16             if(need>left) return false;
17             left-=need;
18             if(w>1){
19                 lazy[i+1]+=need;
20                 if(i+w<=n)
21                     lazy[i+w]-=need;
22             }
23         }
24     }
25     return true;
26 }
27 int main(){
28     while(~scanf("%d%d%d",&n,&m,&w)){
29         int L=1,R=0;
30         for(int i=1;i<=n;i++){
31             scanf("%d",&a[i]);
32             R=max(R,a[i]);
33         }
34         R+=m;
35         while(L<=R){
36             int mid=(L+R)>>1;
37             if(judge(mid))
38                 L=mid+1;
39             else
40                 R=mid-1;
41         }
42         printf("%d
",L-1);
43     }
44     return 0;
45 }

View Code

end

Codeforces Round #262 (Div. 2)

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