UVALive - 3716 DNA Regions

题意：给出两个字符串，求出一个尽量长的序列使得有不超过p%的x满足Ax!=Bx

思路：首先明确条件是

(i-j)*p>=(sum[i]-sum[j])*100 => i*p-sum[i]*100 >= j*p-sum[j]*100,

sum[i]表示前i个的不相等的个数,然后就是构造一个单调递减队列，储存结果,然后二分找出最小的j，第一个入队列的是一定是<0,否则就是目前为止整个都可以

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int MAXN = 150010;

char s1[MAXN],s2[MAXN];
int sum[MAXN];
int n,p;
vector<int> q;
int idx[MAXN];

int main(){
    while (scanf("%d%d",&n,&p) != EOF && n+p){
        scanf("%s%s",s1+1,s2+1);
        sum[0] = 0;
        q.clear();
        q.push_back(0);
        idx[0] = 0;
        int ans = 0;
        for (int i = 1; i <= n; i++){
            sum[i] = sum[i-1] + (s1[i] != s2[i]);
            int t = q.size() - 1;
            if (i*p-sum[i]*100 < q[t]){
                q.push_back(i*p-sum[i]*100);
                idx[q.size()-1] = i;
            }
            int l = 0,r = q.size() - 1;
            int temp = i*p-sum[i]*100;
            while (l < r){
                int mid = (l+r) >> 1;
                if (temp >= q[mid])
                    r = mid;
                else l = mid+1;
            }
            ans = max(ans,i-idx[l]);
        }
        if (ans)
            printf("%d\n",ans);
        else printf("No solution.\n");
    }
    return 0;
}