棰樼洰閾炬帴錛?a style="color: #6a3906; text-decoration: none;">http://poj.org/problem?id=3104榪欓亾棰樻槸鎴戝仛鐨勭涓閬撲簩鍒嗘灇涓劇瓟妗堢殑棰樺惂錛屽彲鏄疄闄呬笂鍋氱殑涔熶笉鎬庝箞鏍?#8230;…
棣栧厛浜屽垎鏋氫婦絳旀錛屼篃灝辨槸鏈鐭椂闂達(dá)紝濡傛灉a[i]<t鐨勮瘽鐩存帴璁╁畠鑷劧椋庡共灝辮錛屽鏋滃ぇ浜巘灝辮綆椾嬌鐢ㄥ嚑嬈￠騫插櫒錛屽亣璁緓1鏄嚜鐒墮騫叉椂闂達(dá)紝x2鏄嬌鐢ㄩ騫插櫒鐨勬鏁幫紝閭ｄ箞x1+x2=t錛寈1+x2*k=a[i]錛屼袱涓柟紼嬭仈绔嬶紝姹傚嚭鏉ヤ嬌鐢ㄧ殑嬈℃暟搴旇鏄?a[i]-t)/(k-1)鐨勪笂鐣岋紝鎵浠ユ妸鎵鏈夌殑浣跨敤鏁ｇ儹鍣ㄥ拰鑷劧椋庡共鐨勬椂闂村姞鍒頒竴璧鳳紝鍜屾灇涓劇殑絳旀姣旇緝錛岀劧鍚庣戶緇簩鍒嗗氨琛屻?br />
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1 #include <cstdio>
2 #include <cmath>
3 int maxt, t, a[100100], k, n;
4 bool check(int tt){
5 double rt = 0;
6 for (int i = 0; i < n; i++)
7 if (a[i] > tt) rt += ceil(double(a[i] - tt) / double(k - 1));
8 if (rt <= tt) return 1;
9 else return 0;
10 }
11 void solve(){
12 int l = 0, r = maxt;
13 while (l <= r){
14 int m = (l + r) >> 1;
15 if (check(m)){
16 if (!check(m - 1)){
17 t = m; return;
18 }
19 r = m - 1;
20 }
21 else l = m + 1;
22 }
23 }
24 int main(){
25 while (~scanf("%d", &n)){
26 maxt = 0;
27 for (int i = 0; i < n; i++){
28 scanf("%d", &a[i]);
29 if (a[i] > maxt) maxt = a[i];
30 }
31 scanf("%d", &k);
32 solve();
33 printf("%d\n", t);
34 }
35 return 0;
36 
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