HDU 1274（集训比赛2B_D题）解题报告

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1274

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题意：输入一个字符串，按题意所要求的进行展开

思路：看到这种具有多次类似操作的展开可以想到递归，从左向右读取字符，遇到（则递归，否则直接输出。递归函数是有返回值的，在不是直接输出字符串的情况，需要递归调用之前的展开次数。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
int n;int m;
const int MAXN =1e5+10;
ll a[MAXN];
int far =0;
ll sum;

int jud(ll mid){
    int man =0;
    ll s=0;
    for(int i=1;i<=far;i++){
        s+=a[i];
        while(s+i>=mid){
            s-=mid-i;
            man++;
        }
    }
    if(s>0) man++;
    //printf("man:%d\n",man);
    if(man>m)    return 1;
    else    return 0;
    
}

ll proce(){
    ll left =far+1;
    ll right =far+sum;
    ll mid =(left+right)/2;
    while(right>=left){
        mid =(left+right)/2;
        //printf("left:%d right:%d mid:%d\n",left,right,mid);
        if(jud(mid)){
            left =mid+1;
        }else{
            right =mid-1;            
        }
    }
    return left;
}

int main(void){
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
        if(a[i]!=0)    far =i;
        sum+=a[i];
    }
    ll ans = proce();
    printf("%lld\n",ans);
    return 0;

}