题干：
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
题目大意：
就是说啊我有一堆放着球的盒子（但是可能有的盒子中没有球），我们现在选择一个有球的盒子，然后把里面的球全都拿出来并且给后面的分球，挨个分球（比如我从2号盒子拿出所有球来了，那么就从3号节点开始分球，，4号，5号这样，如果到头了那就再回到1号盒子开始分球）。现在给你了一共n个盒子，每个盒子在过程结束时的球数，以及是在最后发完球的时候是停在哪个盒子上了（也就是最后一个球是给的哪个盒子）。问你这些盒子中的每个盒子最初由多少球在里面。
解题报告：
这题刚开始就想着模拟，，，结果wa4了。。。其实就是倒着模拟整个过程就好了。
AC代码：
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 ;
}
错误代码：（这个是真的不用看了，，面目全非了）
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];//从0开始读入
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;
//先遍历最小
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]];//先存下来
// a[ans[1]]=0;
// ll up = all / n;//ans[1] == n-1 ?
// all = all-up;//还剩多少需要发出去
// 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 ;
}
总结：
其实这题还可以：
首先不难证明出必须要选择最小的，，，（一开始以为还有可能是第二小的，后来发现题意理解错了，，我出发点的值要变成1，一定是发了完整一圈以后的结果，所以和图形(笛卡尔坐标系?或者一个柱形图)结合一下就显然可证。）然后看在读入的x左方有没有这个最小值，如果有的话，就用最小值的当起点，如果多个，就用最右端的；如果左方没有，那就看右方，如果有的话那就用，如果有多个，那就用最右端的、、、应该这段口胡的算法没错、、、抽空尝试实现一下。（也就是先看左边，如果有就找离x最近的；否则看 右边，此时一定有，因为要保证有解，我们看离x最远的）
