題干：

Little Vasya had?n?boxes with balls in the room. The boxes stood in a row and were numbered with numbers from 1 to?n?from left to right.

Once Vasya chose one of the boxes, let's assume that its number is?i, took all balls out from it (it is guaranteed that this box originally had at least one ball), and began putting balls (one at a time) to the boxes with numbers?i?+?1,?i?+?2,?i?+?3?and so on. If Vasya puts a ball into the box number?n, then the next ball goes to box?1, the next one goes to box?2?and so on. He did it until he had no balls left in his hands. It is possible that Vasya puts multiple balls to the same box, and it is also possible that one or more balls will go to the box number?i. If?i?=?n, Vasya puts the first ball into the box number?1, then the next ball goes to box?2?and so on.

For example, let's suppose that initially Vasya had four boxes, and the first box had?3?balls, the second one had?2, the third one had?5?and the fourth one had?4balls. Then, if?i?=?3, then Vasya will take all five balls out of the third box and put them in the boxes with numbers:?4,?1,?2,?3,?4. After all Vasya's actions the balls will lie in the boxes as follows: in the first box there are?4?balls,?3?in the second one,?1?in the third one and?6?in the fourth one.

At this point Vasya has completely forgotten the original arrangement of the balls in the boxes, but he knows how they are arranged now, and the number?x?— the number of the box, where he put the last of the taken out balls.

He asks you to help to find the initial arrangement of the balls in the boxes.

Input

The first line of the input contains two integers?n?and?x?(2?≤?n?≤?105,?1?≤?x?≤?n), that represent the number of the boxes and the index of the box that got the last ball from Vasya, correspondingly. The second line contains?n?space-separated integers?a1,?a2,?...,?an, where integer?ai?(0?≤?ai?≤?109,?ax?≠?0) represents the number of balls in the box with index?i?after Vasya completes all the actions.

Please, do not use the?%lld?specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the?%I64d?specifier.

Output

Print?n?integers, where the?i-th one represents the number of balls in the box number?i?before Vasya starts acting. Separate the numbers in the output by spaces. If there are multiple correct solutions, you are allowed to print any of them.

Examples

Input

4 4 4 3 1 6

Output

3 2 5 4

Input

5 2 3 2 0 2 7

Output

2 1 4 1 6

Input

3 3 2 3 1

Output

1 2 3

題目大意：

? ?就是說啊我有一堆放著球的盒子（但是可能有的盒子中沒有球），我們現在選擇一個有球的盒子，然后把里面的球全都拿出來并且給后面的分球，挨個分球（比如我從2號盒子拿出所有球來了，那么就從3號節點開始分球，，4號，5號這樣，如果到頭了那就再回到1號盒子開始分球）。現在給你了一共n個盒子，每個盒子在過程結束時的球數，以及是在最后發完球的時候是停在哪個盒子上了（也就是最后一個球是給的哪個盒子）。問你這些盒子中的每個盒子最初由多少球在里面。

解題報告：

? ? 這題剛開始就想著模擬，，，結果wa4了。。。其實就是倒著模擬整個過程就好了。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; ll a[MAX]; int main() {ll n,x,tmp=0;cin>>n>>x;ll minn = 0x3f3f3f3f3f3f;for(int i = 1; i<=n; i++) scanf("%lld",a+i),minn = min(minn,a[i]);for(int i = 1; i<=n; i++) a[i] -= minn;while(a[x] != 0) {a[x]--;x--;tmp++;if(x==0) x=n;}a[x] = tmp + minn*n;for(int i = 1; i<=n; i++) printf("%lld%c",a[i],i == n ? '\n' : ' ');return 0 ;}

?

錯誤代碼：（這個是真的不用看了，，面目全非了）

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; ll a[MAX]; ll n,x; ll ans1[MAX],ans2[MAX]; bool ok(int i) { // ll up = a[i]; // n; // ll tmp = a[i] - up; // if(tmp % n + i == x) return 1; // else return 0;return 1; } int main() {int cnt1=0,cnt2=0;cin>>n>>x;x--;for(int i = 0; i<n; i++) cin>>a[i];//從0開始讀入ll minn = 0x3f3f3f3f3f; for(int i = 0; i<n; i++) {if(a[i] < minn) {cnt1 = 0;ans1[++cnt1] = i;minn = a[i];}else if (a[i] == minn) {ans1[++cnt1] = i;}}ll minn2 = 0x3f3f3f3f3f;for(int i = 0; i<n; i++) {if(a[i] < minn2 && a[i] > minn) {cnt2=0;ans2[++cnt2] = i;minn2 = a[i];}else if(a[i] == minn2) {ans2[++cnt2] == i;}}int that=-1;//先遍歷最小for(int i = 1; i<=cnt1; i++) {if(ok(ans1[i])) {that = ans1[i];break;}}if(that != -1) {ll up = a[that];ll tmp = that <= x ? x-that : x+n-that;a[that]=0;a[that]-=tmp;for(int i = 0; i<n; i++) {if(i != that) a[i] -= up;else a[i] -= up*n;}for(int i = 1; i<=tmp; i++) {a[(that+i)%n]--;} for(int i = 0; i<n; i++) {printf("%lld%c",i == that ? -a[i] : a[i] , i == (n-1) ? '\n' : ' ');}return 0 ;}for(int i = 1; i<=cnt2; i++) {if(ok(ans2[i])) {that = ans2[i];break;}}ll up = a[that];ll tmp = that < x ? x-that : x+n-that;a[that]=0;a[that] -=tmp;for(int i = 0; i<n; i++) {if(i != that) a[i] -= up;else a[i] -= up*n;}for(int i = 1; i<=tmp; i++) {a[(that+i)%n]--;} for(int i = 0; i<n; i++) {printf("%lld%c",i == that ? -a[i] : a[i] , i == (n-1) ? '\n' : ' ');} // if(cnt == 1 && ans[1] <= x) { // ll all = a[ans[1]];//先存下來 // a[ans[1]]=0; // ll up = all / n;//ans[1] == n-1 ? // all = all-up;//還剩多少需要發出去 // for(int i = 0; i<n; i++) { // if(i != ans[1]) a[i] += up; // } // for(int i = 1; i<=all; i++) { // a[(ans[1]+i)%n]++; // } // }return 0 ;}

總結：

? 其實這題還可以：

? ? ? ? ?首先不難證明出必須要選擇最小的，，，（一開始以為還有可能是第二小的，后來發現題意理解錯了，，我出發點的值要變成1，一定是發了完整一圈以后的結果，所以和圖形(笛卡爾坐標系?或者一個柱形圖)結合一下就顯然可證。）然后看在讀入的x左方有沒有這個最小值，如果有的話，就用最小值的當起點，如果多個，就用最右端的；如果左方沒有，那就看右方，如果有的話那就用，如果有多個，那就用最右端的、、、應該這段口胡的算法沒錯、、、抽空嘗試實現一下。（也就是先看左邊，如果有就找離x最近的；否則看 右邊，此時一定有，因為要保證有解，我們看離x最遠的）

總結

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