http://codeforces.com/contest/732/problem/B
题目要求任意两个连续的日子都要 >= k
那么如果a[1] + a[2] < k，就要把a[2]加上数字使得总和 = k，因为这样能传递到a[3]而且是最优的，
a[2]不需要加那么多，只需要加到两个的总和 = k即可，
很坑爹的地方就是 n = 1的时候，其实题目都说了，before the next n days 他们已经走了k
所以n = 1的时候，直接输出0和a[1]即可。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 1000 + 20;
int a[maxn], b[maxn];
void work() {
int n, k;
cin >> n >> k;
for (int i = 1; i <= n; ++i) {
cin >> a[i];
b[i] = a[i];
}
if (n == 1) {
cout << 0 << endl;
cout << a[1] << endl;
return;
}
for (int i = 2; i <= n; ++i) {
if (a[i] + a[i - 1] < k) {
int add = k - (a[i] + a[i - 1]);
a[i] += add;
}
}
int ans = 0;
for (int i = 1; i <= n; ++i) {
ans += a[i] - b[i];
}
cout << ans << endl;
for (int i = 1; i <= n; ++i) {
cout << a[i] << " ";
}
cout << endl;
}
int main() {
#ifdef local
freopen("data.txt","r",stdin);
#endif
work();
return 0;
}
View Code
