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Gym 100851E Easy Problemset (水题,模拟)

题意:给定 n 个裁判,然后每个都一些题目,现在要从每一个按顺序去选出 k 个题,并且这 k 个要按不递减顺序,如果没有,就用50补充。

析:就按他说的来,直接模拟就好。

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 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 LL LNF = 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, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *Hex[] = {"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;}struct node{    int val, id;    node(int v, int i) : val(v), id(i) { }};int a[15][15];int main(){    freopen("easy.in", "r", stdin);    freopen("easy.out", "w", stdout);    int k;    while(scanf("%d %d", &n, &k) == 2){        int x = 0;        memset(a, -1, sizeof a);        for(int i = 0; i < n; ++i){            scanf("%d", &m);            x = Max(x, m);            for(int j = 0; j < m; ++j)                scanf("%d", a[i]+j);        }        int cnt = 0, ans = 0;        for(int i = 0; i < x && cnt < k; ++i){            for(int j = 0; j < n && cnt < k; ++j)                if(a[j][i] >= ans){                    ans += a[j][i];                    ++cnt;                }                else if(a[j][i] == -1) { ans += 50; ++cnt;  break; }        }        while(cnt < k)  ans += 50, ++cnt;        printf("%d\n", ans);    }    return 0;}

 

Gym 100851E Easy Problemset (水题,模拟)