题解归档 - cf2226A
本文最后由方少年更新于2026 年 6 月 28 日,已超过0天没有更新。如果文章内容或图片资源失效,请留言反馈,将会及时处理,谢谢!
题解归档 - cf2226A
本文由 cf-code 本地题解库自动归档;公开内容以本地 AC/验证版本为准。
- 本地编号:
cf2226A - 本地来源:
problems/cf2226A/idea.md - 题目链接:https://codeforces.com/contest/2226/problem/A
- 原始标题:cf2226A - Disturbing Distribution
思路
cf2226A - Disturbing Distribution
pattern: partition into nondecreasing subsequences with positive products.
claim: values greater than 1 are never cheaper to multiply together, because xy >= x+y for x,y>=2. Thus each >1 value can be paid separately. A value 1 costs nothing if it is placed before some later >1 in that operation; all trailing ones after the last >1 need one separate operation of cost 1. If all values are 1, one operation costs 1.
check: samples [1,2,1,2,3] -> 7, [3,2,1] -> 6, all ones -> 1.
代码
来源:problems/cf2226A/solution.cpp
/* Author: likely
* Time: 2026-06-27
**/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll MOD=676767677;
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int t;
cin>>t;
while(t--){
int n,last=-1;
cin>>n;
ll ans=0;
vector<int>a(n+1);
for(int i=1;i<=n;i++){
cin>>a[i];
if(a[i]>1){
ans+=a[i];
last=i;
}
}
if(last==-1){
cout<<1<<"\n";
continue;
}
if(last<n) ans++;
cout<<(ans%MOD)<<"\n";
}
return 0;
}
~ ~ The End ~ ~
文章标题:题解归档 - cf2226A
文章链接:https://www.fangshaonian.cn/archives/239/
最后编辑:2026 年 6 月 28 日 19:05 By 方少年
许可协议: 署名-非商业性使用-相同方式共享 4.0 国际 (CC BY-NC-SA 4.0)