CodeForces 279B Books (滑動窗口)
阿新 • • 發佈:2017-05-13
mat div puts efi freopen 代碼 ble cto tac
題意:給定n本書的閱讀時間,然後你從第 i 本開始閱讀,問你最多能看多少本書在給定時間內。
析:就是一個滑動窗口的水題。
代碼如下:
#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> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 10; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -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 bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int a[maxn]; int main(){ scanf("%d %d", &n, &m); int ans = 0; int i = 0, j = 0, sum = 0; while(j < n){ scanf("%d", a + j); sum += a[j]; if(sum <= m) ans = max(ans, j-i+1); else while(sum > m) sum -= a[i], ++i; ++j; } printf("%d\n", ans); return 0; }
CodeForces 279B Books (滑動窗口)