codefroces 873 B. Balanced Substring && X73(前缀和思想）

In the first example you can choose the substring [3,?6]. It is balanced, and its length is 4. Choosing the substring [2,?5] is also possible.

#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <queue> #include <stack> #include <cstdlib> #include <iomanip> #include <cmath> #include <cassert> #include <ctime> #include <cstdlib> #include <map> #include <set> using namespace std; #pragma comment(linker, "/stck:1024000000,1024000000") #define lowbit(x) (x&(-x)) #define max(x,y) (x>=y?x:y) #define min(x,y) (x<=y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.1415926535897932384626433832 #define ios() ios::sync_with_stdio(true) #define INF 1044266558 #define mem(a) (memset(a,0,sizeof(a))) typedef long long ll; char s[100005]; int a[200005],n,ans=0,val=0; int main() { scanf("%d%s",&n,&s); memset(a,0,sizeof(a)); for(int i=0;i<n;i++) { ans+=s[i]==‘0‘?-1:1; if(ans==0) val=i+1; else { if(!a[ans+n]) a[ans+n]=i+1; else val=max(val,i-a[ans+n]+1);//若中间值存在必为0,因此左右指针对应的值相等 } } printf("%d\n",val); return 0; }

给你一个数组，求数组中连续区间内和为73的区间个数

输入n(0<n<=100000),接下来输入n个数a_i(0<a_i<75);

输出

连续区间内和为73的区间个数，若没有输出"X"

前缀和维护数组。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <cstdlib>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll val[100005],n,ans=0,k=0;
map<ll,ll>a;
int main()
{
    scanf("%lld",&n);
    a[0]=1;
    for(int i=1;i<=n;i++) scanf("%lld",&val[i]);
    for(int i=1;i<=n;i++)
    {
        ans+=val[i];
        a[ans]++;
        if(ans>=73) k+=a[ans-73];
    }
    if(k) printf("%lld\n",k);
    else printf("X\n");
    return 0;
}