D. Prefix Permutation Sums

合集 - 模拟(54)

1.G. Vlad and Trouble at MIT2024-02-202.P2899 [USACO08JAN] Cell Phone Network G2024-02-203.Watering an Array2024-02-234.P2055 [ZJOI2009] 假期的宿舍2024-03-085.T432579 大西洋船王2024-03-086.T432580 星云 hard ver.2024-03-087.P7072 [CSP-J2020] 直播获奖2024-03-098.E - Insert or Erase2024-03-109.D - String Bags2024-03-1010.P2671 [NOIP2015 普及组] 求和2024-03-1011.P1884 [USACO12FEB] Overplanting S2024-03-1012.F. Microcycle2024-03-1213.E. Increasing Subsequences__22024-03-1314.G. Path Prefixes2024-03-1415.F. Rudolf and Imbalance2024-03-1416.P1355 神秘大三角2024-03-2017.[ARC172A] Chocolate2024-03-2518.F. 0, 1, 2, Tree!2024-03-2919.E. Nearly Shortest Repeating Substring2024-03-2920.P1354 房间最短路问题2024-03-2921.P3258 [JLOI2014] 松鼠的新家2024-04-0222.P3052 [USACO12MAR] Cows in a Skyscraper G2024-04-0523.P1337 [JSOI2004] 平衡点 / 吊打XXX2024-04-0724.P3745 [六省联考 2017] 期末考试2024-04-0825.D. Inaccurate Subsequence Search2024-04-0926.E. Long Inversions2024-04-0927.C. Deep Down Below2024-04-0928.C. Rings2024-04-0929.F. Unfair Game2024-04-1030.C. Inversion Graph2024-04-1031.D. Cross Coloring2024-04-1232.D2. Set To Max (Hard Version)2024-04-1233.C. Array Game2024-04-1234.F. Alex's whims2024-04-1235.Queue Sort2024-04-1236.P10252 线性变换2024-04-1637.P10288 [GESP样题 八级] 区间2024-04-1638.P2161 [SHOI2009] 会场预约2024-04-1739.L2-001 紧急救援2024-04-1840.L2-002 链表去重2024-04-1841.L2-013 红色警报2024-04-1842.L2-014 列车调度2024-04-1943.L2-022 重排链表2024-04-1944.L2-023 图着色问题2024-04-1945.L2-039 清点代码库2024-04-1946.L2-034 口罩发放2024-04-1947.L3-037 夺宝大赛2024-04-2348.仓库规划2024-04-2349.因子化简2024-04-2350.C. Torn Lucky Ticket2024-04-2751.俄罗斯方块2024-05-1452.F. Cutting Game2024-05-21

53.D. Prefix Permutation Sums2024-06-17

54.P3304 [SDOI2013] 直径2024-07-15

收起

原题链接

题解

1.缺少一个前缀和，缺少在哪了？
如果缺少在 $i<n$ 的地方，则会出现一个两个数之和，即缺少两个数
否则会只缺少一个数
2.两个数之和可能大于 $n$，也可能不
3.虽然 $a_i$ 达到了 $1e18$ 但是 $n\le 2e5$ ，所以可以用数组记录出现的数

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;

ll pre[200005] = {0};
ll vis[400005] = {0};

inline void read(ll &x) {
	x = 0;
	ll flag = 1;
	char c = getchar();
    while(c < '0' || c > '9'){
        if(c == '-') flag = -1;
        c = getchar();
    }
	while(c >= '0' && c <= '9') {
		x = (x << 3) + (x << 1) + (c ^ 48);
		c = getchar();
	}
	x *= flag;
}

inline void write(ll x) {
    if(x < 0){
    	putchar('-');
		x = -x;
	}
    if(x > 9)
		write(x / 10);
    putchar(x % 10 + '0');
}

void solve()
{
        ll n;
        read(n);

        for(int i=1;i<=2*n;i++) vis[i]=0;
        ll sum = 0;
        int flag=2;
        for(ll i = 1; i < n; i++) {
            read(pre[i]);
            if(pre[i]-pre[i-1]>n+n-1LL)
            {
                flag=0;
            }
            if(!flag) continue;
            if(++vis[pre[i] - pre[i - 1]] == 2 || pre[i] - pre[i - 1] > n) {
                flag--;
                sum = pre[i] - pre[i - 1];
            }
        }

        if(!flag)
        {
            puts("no");
            return;
        }

        if(!sum)
        {
            int cnt=0;
            for(int i=1;i<=n;i++) cnt+=(vis[i]!=0);
            if(cnt==n-1LL) puts("yes");
            else puts("no");
            return ;
        }

        ll tem = sum;
        for(ll i = 1; i <= n; i++) {
            if(!vis[i]) sum -= i;
        }
        if(!sum) puts("YES");
        else puts("NO");
}
int main()
{
    ll t;
    read(t);
    while(t--)
    {
        solve();
    }
    return 0;
}