Codeforces Beta Round #4 (Div. 2 Only)

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A

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;

int main(){
int n;
while(scanf("%d" , &n) != EOF){
if(n < 4)
printf("NO\n");
else{
if(n%2 == 0)
printf("YES\n");
else
printf("NO\n");
}
}

return 0;
}

B

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;

#define MAXN 35

int d , sumTime;
int minTime[MAXN] , maxTime[MAXN];

int main(){
int tmpMin , tmpMax;
while(scanf("%d%d" , &d , &sumTime) != EOF){
tmpMin = tmpMax = 0;
for(int i = 0 ; i < d ; i++){
scanf("%d%d" , &minTime[i] , &maxTime[i]);
tmpMin += minTime[i];
tmpMax += maxTime[i];
}
if(tmpMin > sumTime || sumTime > tmpMax)
printf("NO\n");
else{
int mark = 0;
printf("YES\n");
for(int i = 0 ; i < d ; i++){
if(tmpMin != sumTime){
int count = minTime[i];
for(int j = minTime[i]+1 ; j <= maxTime[i] ; j++){
if(tmpMin == sumTime)
break;
count++;
tmpMin++;
}
if(!mark){
printf("%d" , count);
mark = 1;
}
else
printf(" %d" , count);
}
else{
if(!mark){
printf("%d" , minTime[i]);
mark = 1;
}
else
printf(" %d" , minTime[i]);
}
}
}
}
return 0;
}

C

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define MAXN 100010
#define N 40

int n , pos;
char ch[MAXN]
;
int vis[MAXN];

int main(){
char name
;
while(scanf("%d" , &n) != EOF){
memset(vis , 0 , sizeof(vis));
pos = 0;

int mark , cnt , p;
for(int i = 0 ; i < n ; i++){
scanf("%s" , name);
mark = 1;
for(int j = 0 ; j < pos ; j++){
if(strcmp(ch[j] , name) == 0){
mark = 0;
cnt = vis[j]++;
p = j;
break;
}
}
if(mark){
printf("OK\n");
strcpy(ch[pos] , name);
vis[pos++]++;
}
else
printf("%s%d\n" , ch[p] , cnt);
}
}
return 0;
}

D

/*
思路：dp+路径输出
分析：
1 题目要求的是找到最多的信封的个数并输出编号，如果没有则输出0
2 很明显额矩形嵌套问题，利用dp求解，路径输出利用回溯法记录前驱点。

代码：
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define MAXN 5010

int n , w , h;
int pos , ans , mark;
int pre[MAXN];
int dp[MAXN];
struct envelop{
int w;
int h;
int number;
}e[MAXN];

/*排序*/
bool cmp(envelop e1 , envelop e2){
if(e1.w != e2.w)
return e1.w < e2.w;
return e1.h < e2.h;
}

/*事先判断有没有满足的信封*/
bool judge(){
for(int i = 0 ; i < n ; i++){
if(e[i].w > w && e[i].h > h){
pos = i;
return true;
}
}
return false;
}

/*回溯法输出*/
void output(int x){
if(pre[x] == x){
printf("%d" , e[x].number);
return;
}
output(pre[x]);
printf(" %d" , e[x].number);
}

void solve(){

ans = 0;
for(int i = pos ; i < n ; i++){
dp[i] = 1;
pre[i] = i;
for(int j = pos ; j < i ; j++){
if(e[j].w > w && e[j].h > h && e[i].w > e[j].w && e[i].h > e[j].h && dp[j]+1 > dp[i]){
dp[i] = dp[j] + 1;
pre[i] = j;
}
}
if(ans < dp[i]){
ans = dp[i];
mark = i;
}
}
printf("%d\n" , ans);
output(mark);
printf("\n");
}

int main(){
while(scanf("%d%d%d" , &n , &w , &h) != EOF){
for(int i = 0 ; i < n ; i++){
scanf("%d%d" , &e[i].w , &e[i].h);
e[i].number = i+1;
}
sort(e , e+n , cmp);
if(!judge())
printf("0\n");
else
solve();
}
return 0;
}