CF--Technocup 2017 - Elimination Round 2(ABCDE)
2016-11-22 21:29
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真的不是模拟专场么?(大雾
n*k是4e10,肯定超时,这时因为需要最少的钱数,想到二分,二分要保证钱数多的那个如果不可以,那么钱数少的就绝对不可以,这个问题保证的话就需要对输入做些手脚,按钱升序,钱同容量降序排序后,每个钱数只留下容量最大的(贪心?),且如果钱数多的容量反而小了,就不要了
最后判断的时候我加了一个小优化,估计没有也不会t,就是存下最长那段路的距离,如果这个能跑下来,那每一段都可以,如果不能,直接gg
最后算一下时间和,和t比一下就好了。。
t = 2*s[i] - min(V-s[i],s[i]);
除了祖先以外的所有0必须要改,所有大于n-1的所有要改,剩下的数字需要是连续的,中间不能断,出现断的先从上面必须要改的改,然后如果不够,就从剩下的中最大的改,去掉叶子总不会错
A题:
题意:
遇到ogo换成*** 后面紧接着的所有go忽略,其他字符打印出来
tip:
while大法好#include <cstdio> #include <iostream> #include <cstring> using namespace std; const int maxn = 110; int n; char s[maxn]; void init(){ scanf("%s",s); for(int i = 0 ; i < n ; i++){ // printf("sf %c\n",s[i]); if(s[i] == 'o' && s[i+1] == 'g' && s[i+2] == 'o'){ printf("***"); i += 2; while(i + 2 < n && s[i+1] == 'g' && s[i+2] == 'o') i += 2; } else printf("%c",s[i]); } printf("\n"); } int main(){ while(~scanf("%d",&n)){ init(); } }
B题:
题意:
舞台上摆放灯光,灯光所在位置不能有人,四个方向的射线需要有人tip:
模拟一下。。不要太暴力就好#include <cstdio> #include <iostream> #include <cstring> using namespace std; const int maxn = 1010; typedef pair<int,int> pii; int dir[4][2] = { {0,1},{0,-1},{-1,0},{1,0}}; int n,m,a[maxn][maxn],tot = 0,rf[maxn],rr[maxn],cu[maxn],cd[maxn]; pii ans[maxn*maxn]; void init(){ tot = 0; memset(rf,-1,sizeof(rf)); memset(rr,-1,sizeof(rf)); memset(cu,-1,sizeof(rf)); memset(cd,-1,sizeof(rf)); for(int i = 0; i < n ;i++) for(int j = 0 ; j < m ;j++){ scanf("%d",&a[i][j]); if(a[i][j] == 0) ans[tot++] = make_pair(i,j); else{ if(rf[i] == -1){ rf[i] = j;//右边的都可以开左灯 } rr[i] = j;//左边的都可以开右灯 if(cu[j] == -1){ cu[j] = i;//下边的都可以开上灯 } cd[j] = i;//上边的都可以开下灯 } } } void sov(){ int ak = 0; for(int i = 0 ;i < tot ;i++){ int x = ans[i].first,y = ans[i].second; if(y > rf[x]&&rf[x]!=-1) ak++; if(y < rr[x]&&rr[x]!=-1) ak++; if(x > cu[y]&&cu[y]!=-1) ak++; if(x < cd[y]&&cd[y]!=-1) ak++; } printf("%d\n",ak); } int main(){ while(~scanf("%d%d",&n,&m)){ init(); sov(); } }
C题:
题意:
多辆车,起始满油,整个过程中有k个加油站,在每个加油站加油不需要钱和时间,整个过程需要的钱就是车的钱,其实就是问你选哪个车,其中每个车的邮箱不容量不同,2分钟1km需1L油,1分钟1km需2L油。问在t时间内能走s米的最小花费。tip:
个人认为比D难模拟。正常情况下x米就需要2x分钟,需要xL油,但是因为还有油就可以用油换时间,每多1L油可以-1s,减到x分钟为止(最快x分钟),n*k是4e10,肯定超时,这时因为需要最少的钱数,想到二分,二分要保证钱数多的那个如果不可以,那么钱数少的就绝对不可以,这个问题保证的话就需要对输入做些手脚,按钱升序,钱同容量降序排序后,每个钱数只留下容量最大的(贪心?),且如果钱数多的容量反而小了,就不要了
for(int i = 1 ;i < n ; i++){ if(tmp[i].first == tmp[i-1].first) continue; else if(tmp[i].second <= a[pp-1].second) continue; else a[pp++] = tmp[i]; }
最后判断的时候我加了一个小优化,估计没有也不会t,就是存下最长那段路的距离,如果这个能跑下来,那每一段都可以,如果不能,直接gg
最后算一下时间和,和t比一下就好了。。
t = 2*s[i] - min(V-s[i],s[i]);
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int n ,k,s,t,ans,pp;
const int maxn = 2*1e5+10;
typedef pair <int,int> pii ;
pii a[maxn],tmp[maxn];
int m[maxn],aa[maxn];
bool cmp(pii a,pii b){
if(a.first == b.first)
return a.second > b.second;
return a.first < b.first;
}
bool judge(int mid){
int nt = 0;
if(ans > a[mid].second) return false;
for(int i = 0 ; i <= k ;i++){
nt += 2*m[i]-min(a[mid].second-m[i],m[i]);
if(nt >t) return false;
}
if(nt > t) return false;
return true;
}
void sov(){
int l = 0,r = pp-1,all = -1;
while(l <= r){
int mid = l+(r-l)/2;//printf("a[%d].se = %d\n",mid,a[mid].second);
if(judge(mid)) {
r = mid-1;
all = a[mid].first;
}
else l = mid+1;
}
printf("%d\n",all);
}
void init(){
int sum = 0;
ans = -1,pp = 1;
for(int i = 0 ; i < n ; i++)
scanf("%d%d",&tmp[i].first,&tmp[i].second);
sort(tmp,tmp+n,cmp);
a[0] = tmp[0];
for(int i = 1 ;i < n ; i++){ if(tmp[i].first == tmp[i-1].first) continue; else if(tmp[i].second <= a[pp-1].second) continue; else a[pp++] = tmp[i]; }
for(int i = 0 ;i < k ;i++)
scanf("%d",&aa[i]);
sort(aa,aa+k);
m[0] = aa[0];
for(int i = 1 ; i < k ; i++){
m[i] = aa[i]-aa[i-1];
ans = max(ans,m[i]);
}
m[k] = s-aa[k-1];
ans = max(ans,m[k]);
}
int main(){
while(~scanf("%d%d%d%d",&n,&k,&s,&t)){
init();
sov();
}
}
/*
2 2 10 15
10 4
20 6
5 3
*/
D题:
题意:
a搜船,每个b这么长,01串,1表示没有船在那,0表示不知道,问你在剩下的0中至少打哪些位置,能保证至少打到了一艘船tip:
写错了好几次,首先一定打到所有可能位置-b+1就一定能打到船,然后所有可能位置我想了几种模拟,都不太对,最后知道了,两个1中间的所有0,每b个一放入,一定是只多不少,也就把所有位置包括了。。。#include <cstdio> int N, A, B, K, count; char b[200005]; int op[200005]; int main() { scanf("%d %d %d %d %s", &N, &A, &B, &K, b); int s = 0; for (int i = 0; i < N; i++) if (b[i] == '1') { for (int j = s+B-1; j < i; j += B) op[count++] = j; s = i+1; } for (int j = s+B-1; j < N; j += B) op[count++] = j; printf("%d\n", count-A+1); for (int i = 0; i < count-A+1; i++) printf("%d ", op[i]+1); printf("\n"); }
E题:
题意:
树上每个节点都写着他的所有祖先的数量,问最少有多少个写错了(祖先位置固定)tip:
谁告诉我要数据结构写树的?除了祖先以外的所有0必须要改,所有大于n-1的所有要改,剩下的数字需要是连续的,中间不能断,出现断的先从上面必须要改的改,然后如果不够,就从剩下的中最大的改,去掉叶子总不会错
#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> using namespace std; const int maxn = 2*1e5+10; int n,s,ans,tot,k; int a[maxn],b[maxn],c[maxn]; void sov(){ for(int i = 1; i <= c[tot] ;i++){ if(b[i] == 0&&k) k--; else if(b[i] == 0 && k == 0){ b[c[tot]]--; ans++; if(b[c[tot]] == 0) tot--; } } printf("%d\n",ans); } void init(){ ans = 0,tot = 0; int num0 = 0 ,numb = 0; memset(b,0,sizeof(b)); for(int i = 1 ; i <= n ; i++){ scanf("%d",&a[i]); if(i == s) continue; if(a[i] == 0) num0++; else if(a[i] >= n) numb++; else { if(b[a[i]] == 0){ c[tot++] = a[i]; } b[a[i]] ++; } } ans = num0+numb; k = ans; if(a[s]!=0) ans++; a[s] = 0; sort(c,c+tot); tot--; } int main(){ while(~scanf("%d%d",&n,&s)){ init(); sov(); } }
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