CodeForces - 731B(贪心)
2018-02-26 17:20
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题目链接:点击打开链接题目大意:有N天需要买披萨,两种方式,一种是一天一次买两个披萨,另一种是今天买一个明天买一个披萨,问是否能把披萨正好全买了。 解题思路:直接模拟,如果当天所需披萨数是偶数,直接就全买过来,如果是奇数,就只好要明天买一个了。(P.S. 如果明天只有0个,今天是奇数,那就凉了..(明天的披萨数会小于0)需要特判一下别被坑) 以下是代码:#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int a[200005];
int main(){
int n,sum,i;
while(~scanf("%d",&n)){
for(i=0;i<n;i++){
scanf("%d",&a[i]);
}
for(i=0;i<n;i++){
if(a[i]<0){
break;
}
if(a[i]%2==1){
a[i+1]--;
}
}
if(i!=n){
printf("NO\n");
}
else{
if(a[n-1]%2==0){
printf("YES\n");
}
else{
printf("NO\n");
}
}
}
return 0;
}
#include<cstdio>
#include<algorithm>
using namespace std;
int a[200005];
int main(){
int n,sum,i;
while(~scanf("%d",&n)){
for(i=0;i<n;i++){
scanf("%d",&a[i]);
}
for(i=0;i<n;i++){
if(a[i]<0){
break;
}
if(a[i]%2==1){
a[i+1]--;
}
}
if(i!=n){
printf("NO\n");
}
else{
if(a[n-1]%2==0){
printf("YES\n");
}
else{
printf("NO\n");
}
}
}
return 0;
}
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