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  • CodeForces 732D Exams (二分)

    题意:某人要考试,有n天考m个科目,然后有m个科目要考试的时间和要复习多少天才能做,问你他最早考完所有科目是什么时间。

    析:二分答案,然后在判断时,直接就是倒着判,很明显后出来的优先,也就是一个栈。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #define debug() puts("++++");
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 1e5 + 5;
    const int mod = 1e9 + 7;
    const int dr[] = {-1, 1, 0, 0};
    const int dc[] = {0, 0, 1, -1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline int Min(int a, int b){ return a < b ? a : b; }
    inline int Max(int a, int b){ return a > b ? a : b; }
    inline LL Min(LL a, LL b){ return a < b ? a : b; }
    inline LL Max(LL a, LL b){ return a > b ? a : b; }
    inline bool is_in(int r, int c){
        return r >= 0 && r < n && c >= 0 && c < m;
    }
    int a[maxn], b[maxn];
    int ok[maxn], cnt[maxn];
    
    bool judge(int mid){
        memset(ok, -1, sizeof ok);
        int idx = 0, num = 0;
        for(int i = mid; i > 0; --i){
            if(a[i] && ok[a[i]] == -1)  cnt[++idx] = a[i], ++ok[a[i]], ++num;
            else{
                while(idx && !b[cnt[idx]])  --idx;
                if(!idx)  continue;
                ++ok[cnt[idx]];
                if(ok[cnt[idx]] == b[cnt[idx]]) --idx;
            }
        }
    
        return num == m && !idx;
    }
    
    int solve(){
        int l = m, r = n+1;
        while(l < r){
            int mid = (l + r) >> 1;
            if(judge(mid))  r = mid;
            else l = mid + 1;
        }
    
        return l == n+1 ? -1 : l;
    }
    
    int main(){
        while(scanf("%d %d", &n, &m) == 2){
            for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
            a[n+1] = 0;
            b[0] = -1;
            for(int i = 1; i <= m; ++i)  scanf("%d", b+i);
            printf("%d
    ", solve());
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/6263506.html
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