Codeforces 21C Stripe 2 卧槽,出题人母语绝对不是English啊
2014-11-06 20:29
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sum = sigma num[i] (1 <= i <= n)。
s[i] = sigam num[j] (1<= j <= i)。
找到所有满足s[i]*2 =sum-s[i] ,s[i] = 2*(sum-s[i])。然后枚举累加一下。
卧槽,可是题意里明明说每个 piece 里 都要有positive interge。
可是去了这个限制就过了啊。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <ctime>
#include <iomanip>
#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-6)
#define _LL long long
#define ULL unsigned long long
#define LL __int64
#define INF 0x3f3f3f3f
#define Mod 1000000007
/** I/O Accelerator Interface .. **/
#define g (c=getchar())
#define d isdigit(g)
#define p x=x*10+c-'0'
#define n x=x*10+'0'-c
#define pp l/=10,p
#define nn l/=10,n
template<class T> inline T& RD(T &x)
{
char c;
while(!d);
x=c-'0';
while(d)p;
return x;
}
template<class T> inline T& RDD(T &x)
{
char c;
while(g,c!='-'&&!isdigit(c));
if (c=='-')
{
x='0'-g;
while(d)n;
}
else
{
x=c-'0';
while(d)p;
}
return x;
}
inline double& RF(double &x) //scanf("%lf", &x);
{
char c;
while(g,c!='-'&&c!='.'&&!isdigit(c));
if(c=='-')if(g=='.')
{
x=0;
double l=1;
while(d)nn;
x*=l;
}
else
{
x='0'-c;
while(d)n;
if(c=='.')
{
double l=1;
while(d)nn;
x*=l;
}
}
else if(c=='.')
{
x=0;
double l=1;
while(d)pp;
x*=l;
}
else
{
x=c-'0';
while(d)p;
if(c=='.')
{
double l=1;
while(d)pp;
x*=l;
}
}
return x;
}
#undef nn
#undef pp
#undef n
#undef p
#undef d
#undef g
using namespace std;
LL num[100010];
LL ans[2][100010] = {{0},{0}};
LL sum[2][100010] = {{0},{0}};
LL dis[100010] = {0};
int mark[2][100010] = {{0},{0}};
int main()
{
int n;
scanf("%d",&n);
int i;
for(i = 1;i <= n; ++i)
scanf("%I64d",&num[i]);
for(i = 1;i <= n; ++i)
sum[0][i] = sum[0][i-1] + num[i];
for(i = n;i >= 1; --i)
sum[1][i] = sum[1][i+1] + num[i];
for(i = 1;i <= n; ++i)
ans[0][i] = ans[0][i-1] + (num[i] >= 0 ? 1 : 0);
for(i = n;i >= 1; --i)
ans[1][i] = ans[1][i+1] + (num[i] >= 0 ? 1 : 0);
LL S = sum[0]
;
for(i = 1;i <= n; ++i)
if(sum[0][i]*2 == S-sum[0][i] )
mark[0][i] = 1;
for(i = n;i >= 1; --i)
if(sum[1][i]*2 == S-sum[1][i])
mark[1][i] = 1;
for(i = 1;i <= n; ++i)
dis[i] = dis[i-1] + mark[0][i];
LL anw = 0;
for(i = 2;i <= n; ++i)
if(mark[1][i])
anw += dis[i-2];
printf("%I64d\n",anw);
return 0;
}
s[i] = sigam num[j] (1<= j <= i)。
找到所有满足s[i]*2 =sum-s[i] ,s[i] = 2*(sum-s[i])。然后枚举累加一下。
卧槽,可是题意里明明说每个 piece 里 都要有positive interge。
可是去了这个限制就过了啊。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <ctime>
#include <iomanip>
#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-6)
#define _LL long long
#define ULL unsigned long long
#define LL __int64
#define INF 0x3f3f3f3f
#define Mod 1000000007
/** I/O Accelerator Interface .. **/
#define g (c=getchar())
#define d isdigit(g)
#define p x=x*10+c-'0'
#define n x=x*10+'0'-c
#define pp l/=10,p
#define nn l/=10,n
template<class T> inline T& RD(T &x)
{
char c;
while(!d);
x=c-'0';
while(d)p;
return x;
}
template<class T> inline T& RDD(T &x)
{
char c;
while(g,c!='-'&&!isdigit(c));
if (c=='-')
{
x='0'-g;
while(d)n;
}
else
{
x=c-'0';
while(d)p;
}
return x;
}
inline double& RF(double &x) //scanf("%lf", &x);
{
char c;
while(g,c!='-'&&c!='.'&&!isdigit(c));
if(c=='-')if(g=='.')
{
x=0;
double l=1;
while(d)nn;
x*=l;
}
else
{
x='0'-c;
while(d)n;
if(c=='.')
{
double l=1;
while(d)nn;
x*=l;
}
}
else if(c=='.')
{
x=0;
double l=1;
while(d)pp;
x*=l;
}
else
{
x=c-'0';
while(d)p;
if(c=='.')
{
double l=1;
while(d)pp;
x*=l;
}
}
return x;
}
#undef nn
#undef pp
#undef n
#undef p
#undef d
#undef g
using namespace std;
LL num[100010];
LL ans[2][100010] = {{0},{0}};
LL sum[2][100010] = {{0},{0}};
LL dis[100010] = {0};
int mark[2][100010] = {{0},{0}};
int main()
{
int n;
scanf("%d",&n);
int i;
for(i = 1;i <= n; ++i)
scanf("%I64d",&num[i]);
for(i = 1;i <= n; ++i)
sum[0][i] = sum[0][i-1] + num[i];
for(i = n;i >= 1; --i)
sum[1][i] = sum[1][i+1] + num[i];
for(i = 1;i <= n; ++i)
ans[0][i] = ans[0][i-1] + (num[i] >= 0 ? 1 : 0);
for(i = n;i >= 1; --i)
ans[1][i] = ans[1][i+1] + (num[i] >= 0 ? 1 : 0);
LL S = sum[0]
;
for(i = 1;i <= n; ++i)
if(sum[0][i]*2 == S-sum[0][i] )
mark[0][i] = 1;
for(i = n;i >= 1; --i)
if(sum[1][i]*2 == S-sum[1][i])
mark[1][i] = 1;
for(i = 1;i <= n; ++i)
dis[i] = dis[i-1] + mark[0][i];
LL anw = 0;
for(i = 2;i <= n; ++i)
if(mark[1][i])
anw += dis[i-2];
printf("%I64d\n",anw);
return 0;
}
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