标签：follow bit spec class his put nbsp art osi
The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let‘s enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all the houses with larger numbers. In other words, a house is luxurious if the number of floors in it is strictly greater than in all the houses, which are located to the right from it. In this task it is assumed that the heights of floors in the houses are the same.
The new architect is interested in n questions, i-th of them is about the following: "how many floors should be added to the i-th house to make it luxurious?" (for all i from 1 to n, inclusive). You need to help him cope with this task.
Note that all these questions are independent from each other — the answer to the question for house i does not affect other answers (i.e., the floors to the houses are not actually added).
The first line of the input contains a single number n (1?≤?n?≤?105) — the number of houses in the capital of Berland.
The second line contains n space-separated positive integers hi (1?≤?hi?≤?109), where hi equals the number of floors in the i-th house.
Print n integers a1,?a2,?...,?an, where number ai is the number of floors that need to be added to the house number i to make it luxurious. If the house is already luxurious and nothing needs to be added to it, then ai should be equal to zero.
All houses are numbered from left to right, starting from one.
5
1 2 3 1 2
3 2 0 2 0
4
3 2 1 4
2 3 4 0
dp[ i ] 表示从 I 位置到尾的区间最大值；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == ‘-‘) f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int n;
int h[maxn];
int a[maxn];
int dp[maxn];
int main() {
//ios::sync_with_stdio(0);
rdint(n);
for (int i = 1; i <= n; i++)rdint(h[i]);
dp[n] = h[n]; int maxx = h[n];
for (int i = n; i >= 1; i--) {
maxx = max(maxx, h[i]);
dp[i] = maxx;
// cout << dp[i] << endl;
}
for (int i = 1; i <= n; i++) {
if (i == n) {
cout << 0 << endl; return 0;
}
if (h[i] > dp[i+1])cout << 0 << ‘ ‘;
else {
cout << dp[i+1] - h[i] + 1 << ‘ ‘;
}
}
return 0;
}
标签：follow bit spec class his put nbsp art osi
原文地址：https://www.cnblogs.com/zxyqzy/p/10293796.html
