Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals)-831C Jury Marks(思维)
2017-07-14 23:59
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题意:选手有一个初始分数,然后有k个评委的评分,这些分数将依次加到选手的分数上,不过选手只记得在这个过程中的某个时刻n(n<=k)个不同的自己当时的分数(顺序不定),问初始分数有多少种可能?
思路:思维题+STL。首先可以分析将k个评分处理一个前缀和,发现前缀和相同的位置在两个时刻的分数是相同的,然后通过unique函数进行去重。然后会发现题目中突出显示的distinct,给定的n个所知的时刻分数是不同的。所以我们要构造一个初始值来满足存在这n个值都存在的话,只能从这去重之后的cnt个数中进行构造。所以枚举这cnt个不同的前缀和,第二层枚举n个数,从而确定一个初始值,然后再通过初始值判断第三层n个数是否都能通过这前cnt个前缀和得到即可。
Code:
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思路:思维题+STL。首先可以分析将k个评分处理一个前缀和,发现前缀和相同的位置在两个时刻的分数是相同的,然后通过unique函数进行去重。然后会发现题目中突出显示的distinct,给定的n个所知的时刻分数是不同的。所以我们要构造一个初始值来满足存在这n个值都存在的话,只能从这去重之后的cnt个数中进行构造。所以枚举这cnt个不同的前缀和,第二层枚举n个数,从而确定一个初始值,然后再通过初始值判断第三层n个数是否都能通过这前cnt个前缀和得到即可。
Code:
#include <bits/stdc++.h> using namespace std; const int maxn = 2005; const int N = 8e6; int a[maxn], b[maxn], pre[maxn]; int vis[2*N+5]; int k, n, base, ans, cnt, l; map<int, int> mp1; int main() { scanf("%d %d", &k, &n); for(int i = 1; i <= k; ++i) scanf("%d", &a[i]); for(int i = 1; i <= n; ++i) scanf("%d", &b[i]); pre[0] = 0; ans = 0; for(int i = 1; i <= k; ++i) pre[i] = pre[i-1]+a[i]; sort(pre+1, pre+k+1); cnt = unique(pre+1, pre+k+1)-pre; for(int i = 1; i < cnt; ++i) ++mp1[pre[i]]; for(int i = 1; i < cnt; ++i) { for(int j = 1; j <= n; ++j) { base = b[j]-pre[i]; if(vis[base+N]) continue; //每个base值增加一个ans vis[base+N] = 1; for(l = 1; l <= n; ++l) { if(mp1.find(b[l]-base) == mp1.end()) break; //break小优化 } if(l == n+1) ++ans; } } printf("%d\n", ans); return 0; }
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