【2017 Multi-University Training Contest - Team 5】Rikka with Subset
2017-10-04 18:44
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【Link】:
【Description】
给你a数组的n个数的所有2^n个子集的2^n个子集元素的和;
子集元素的和最大为m;
告诉你各个子集元素的和出现的次数;
如
1 2
则0出现1次,1出现1次,2出现一次,3出现一次;
分别对应{},{1},{2},{1,2};
问你能不能复原出原数组;
输出最小字典序的数组;
【Solution】
给你一个b数组;
实际上和背包方案数问题的f数组是一样的;
f[i]表示物品组成的体积为i时的方案数
for (int i = 1;i <= n;i++) for (int j = m ; j >= w[i];j--) if (f[j-w[i]]) f[j] += f[j-w[i]];
我们可以保证,最小的出现过的正数,一定是a[1];
(假设a数组是从小到大的顺序给出);
那么,我们把这个a[1]取走,让其只剩下n-1个物品,使得b数组里面不包含这个a[1]物品的信息;这样的话,就能继续按照”取最小的出现过的正数”这个原则找a[2]了;
使得b数组里面不包含a[1]的信息
逆序做一遍背包就可以了;
按照上面的更新顺序,逆着写一下即可.
【NumberOf WA】
2
【Reviw】
一开始忘记是f[j]+f[j-w[i]]了
【Code】
#include <bits/stdc++.h> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define LL long long #define rep1(i,a,b) for (int i = a;i <= b;i++) #define rep2(i,a,b) for (int i = a;i >= b;i--) #define mp make_pair #define pb push_back #define fi first #define se second #define ms(x,y) memset(x,y,sizeof x) #define ri(x) scanf("%d",&x) #define rl(x) scanf("%lld",&x) #define rs(x) scanf("%s",x+1) #define oi(x) printf("%d",x) #define ol(x) printf("%lld",x) #define oc putchar(' ') #define os(x) printf(x) #define all(x) x.begin(),x.end() #define Open() freopen("F:\\rush.txt","r",stdin) #define Close() ios::sync_with_stdio(0) typedef pair<int,int> pii; typedef pair<LL,LL> pll; const int dx[9] = {0,1,-1,0,0,-1,-1,1,1}; const int dy[9] = {0,0,0,-1,1,-1,1,-1,1}; const double pi = acos(-1.0); const int N = 110; LL n,m; int main(){ //Open(); //Close(); int T; ri(T); while (T--){ rl(n),rl(m); LL ans; //special? if ( m <= n - 1){ ans = (m*(m-1)*2 + m*2) + (n*(n-1) -m*(m-1) - m*2 )*n; }else{ LL temp = m-(n-1); LL temp1 = (n-1)*(n-2)/2; if (temp >= temp1){ ans = n*(n-1); }else{ LL delta = temp1-temp; ans = (n-1)*2 + delta*2*2 + temp*2; } } ol(ans);puts(""); } return 0; }
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