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; }