2017 ACM Arabella Collegiate Programming Contest A. Sherlock Bones GYM101350A

最新推荐文章于 2019-04-06 15:16:24 发布

菜得不行的sb

最新推荐文章于 2019-04-06 15:16:24 发布

阅读量518

点赞数 1

CC 4.0 BY-SA版权

分类专栏：思维

本文链接：https://blog.youkuaiyun.com/cys460714380/article/details/78245588

思维专栏收录该内容

24 篇文章

订阅专栏

本文介绍了一种通过计算区间内奇数位数来解决特定问题的算法。该算法首先统计区间内的奇数位数，然后减去只包含单一奇数的情况以避免重复计算。文章通过C++代码详细展示了算法实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

把问题转换为求区间内1为奇数的区间有多少个
那么直接记录从1开始为奇数和偶数的区间有多少就算出这个了

但是有不合法的就是只有1个1的区间
所以再减一下重复的再加回来

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ans() printf("%d",ans)
#define ansn() printf("%d\n",ans)
#define anss() printf("%d ",ans)
#define lans() printf("%lld",ans)
#define lanss() printf("%lld ",ans)
#define lansn() printf("%lld\n",ans)
#define fansn() printf("%.6f\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define rsz(i,v) for(int i=0;i<(int)v.size();++i)
#define szz(x) ((int)x.size())
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pli pair<ll,int>
#define pll pair<ll,ll>
#define mp make_pair
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
#define sqr(a) ((a)*(a))
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000000+7;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
const ll infl = 10000000000000000;
const int maxn=  200000+10;
const int maxm = 40000+10;
//Pretests passed
int in(int &ret)
{
    char c;
    int sgn ;
    if(c=getchar(),c==EOF)return -1;
    while(c!='-'&&(c<'0'||c>'9'))c=getchar();
    sgn = (c=='-')?-1:1;
    ret = (c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');
    ret *=sgn;
    return 1;
}

char s[maxn];

int main()
{
#ifdef LOCAL
    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCAL

    int t;
    sd(t);
    while(t--)
    {
        int n;
        sd(n);
        ss(s+1);
        ll x[2];
        ll ans = 0;
        int now = 0;
        x[0] = 1;
        x[1] = 0;
        r1(i,n)
        {
            if(s[i]-'0')now^=1;
            ans += x[now^1];
            x[now]++;
        }
        int last = -1;
        for(int i=1;i<=n;++i)
        {
            if(s[i]-'0')
            {
                if(last==-1)last = i;
                else
                {
                    ans -= i - last ;
                    last = i;
                }
            }
        }
        if(last!=-1&&last!=n)ans -= n + 1 - last;
        else if(last==n)--ans;
        last = -1;
        for(int i=n;i>=1;--i)
        {
            if(s[i]-'0')
            {
                if(last==-1)last = i;
                else
                {
                    ans -= last - i ;
                    last = i;
                }
            }
        }
        if(last!=-1&&last!=1)ans -= last ;
        else if(last==1)ans--;
        r1(i,n)if(s[i]-'0')++ans;
        lansn();

    }
    return 0;
}