[POI2011]LIZ-Lollipop（前缀和）

原题

题面

Solution

我是不会告诉你我一开始写了一个二分答案+rand的算法的。。。
这样子你可以获得80分的好成绩！！！

#pragma GCC diagnostic error "-std=c++11"
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<queue>
#define ll long long
#define file(a) freopen(a".in","r",stdin)//;freopen(a".out","w",stdout)
using namespace std;
inline int gi(){
    int sum=0,f=1;char ch=getchar();
    while(ch>'9' || ch<'0'){if(ch=='-')f=-f;ch=getchar();}
    while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
    return f*sum;
}
inline ll gl(){
    ll sum=0,f=1;char ch=getchar();
    while(ch>'9' || ch<'0'){if(ch=='-')f=-f;ch=getchar();}
    while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
    return f*sum;
}
int a[2000010],sum[2000010];
struct node{
    int id,val;
    bool operator<(const node b)const{
        return val<b.val;
    }
}d[2000010];
int main(){
    srand(time(NULL));
#ifndef ONLINE_JUDGE
    file("example");
#endif
    int i,j,n,m,k;
    n=gi();m=gi();
    for(i=1;i<=n;i++){
        char ch=getchar();
        a[i]=ch=='W'?1:2;
    }
    for(i=1;i<=n;i++)sum[i]=sum[i-1]+a[i];
    for(i=1;i<=n;i++){
        d[i]=(node){i,rand()};
    }
    sort(d+1,d+n+1);
    while(m--){
        k=gi();int flag=0;
        for(int x=1;x<=n;x++){
            i=d[x].id;
            int l=i,r=n,ans=0;
            while(l<=r){
                int mid=(l+r)>>1;
                if(sum[mid]-sum[i-1]>k)r=mid-1;
                else if(sum[mid]-sum[i-1]<k)l=mid+1;
                else{ans=mid;break;}
            }
            if(ans){
                printf("%d %d
",i,ans);flag=1;
                break;
            }
        }
        if(!flag)puts("NIE");
    }
    return 0;
}

这是一道比较有思维难度的题目，我们考虑一下如果一段区间有1，那么显然都可以完成对吧。
这里给一个比较简单的证明:
(Sum)表示前i个数的和,(sum_i)表示i往后有多少个T.我们考虑当前这个数如果是T(2),那么有如下两种情况

• i向后延续的长度>1向后延续的长度,则有:
  (Sum-1=a_{sum_1+2}+a_{sum_1+3}+...+a_{sum_1+i})

这个可以这么想,我们把它差分一下,变成:
(a_1+a_2+...+a_{sum_1+i}-a_1-a_2-...-a_{sum_1+1})
因为(1~sum_1)都是2,所以前面的数都是2.
然后我们考虑第一个式子(a_1+a_2+a_3+...+a_{sum_1+i})
因为i向后延续的长度>1向后延续的长度,所以我们考虑把这i个数往前移,然后把1往后延续的长度放到后面,就有
原式=(a_1+a_2+...+a_{sum_1+i}-a_1-a_2-a_{sum_1+1}=Sum-1)

---

• 如果i延续的长度不能够到n,则有:
  (Sum-1=a_{sum_i+1}+a_{sum_i+2}+...+a_{sum_i+i})

同理可证

---

那么就可以写出下面这段代码了.

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<queue>
#define ll long long
#define file(a) freopen(a".in","r",stdin)//;freopen(a".out","w",stdout)
using namespace std;
inline int gi(){
    int sum=0,f=1;char ch=getchar();
    while(ch>'9' || ch<'0'){if(ch=='-')f=-f;ch=getchar();}
    while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
    return f*sum;
}
inline ll gl(){
    ll sum=0,f=1;char ch=getchar();
    while(ch>'9' || ch<'0'){if(ch=='-')f=-f;ch=getchar();}
    while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
    return f*sum;
}
int a[2000010],sum[2000010],Sum,l[2000010],r[2000010];
int main(){
#ifndef ONLINE_JUDGE
    file("example");
#endif
    int i,j,n,m,k;
    n=gi();m=gi();
    for(i=1;i<=n;i++){
        char ch=getchar();
        if(ch=='T')a[i]=2;
        else a[i]=1;
    }
    for(i=n;i;i--)sum[i]=a[i]==2?(sum[i+1]+1):0;
    int Sum=0;
    for(i=1;i<=n;i++){
        Sum+=a[i];
        l[Sum]=1;r[Sum]=i;
        if(a[i]==2)
            if(sum[i]>sum[1])l[Sum-1]=sum[1]+2,r[Sum-1]=sum[1]+i;
            else if(sum[i]!=n-i+1)l[Sum-1]=sum[i]+1,r[Sum-1]=sum[i]+i;
    }
    while(m--){
        k=gi();
        if(l[k])printf("%d %d
",l[k],r[k]);
        else puts("NIE");
    }
    return 0;
}