CodeForces 732B Cormen — The Best Friend Of a Man (贪心)

题意：给定n和k表示，狗要在任意连续两天散步次数要至少为k，然后就是n个数，表示每天的时间，让你增加最少次数使得这个条件成立。

析：贪心，策略是从开始到最后暴力，每次和前面一个相比，如果相加不够k，那么就给当前加上差。

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 double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 500 + 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];

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        for(int i = 0; i < n; ++i)  scanf("%d", a+i);
        int ans = 0;
        for(int i = 1; i < n; ++i){
            ans += Max(0, m - a[i] - a[i-1]);
            a[i] = Max(a[i], m - a[i-1]);
        }
        printf("%d\n", ans);
        for(int i = 0; i < n; ++i){
            if(i) putchar(' ');
            printf("%d", a[i]);
        }
        printf("\n");
    }
    return 0;
}