Description
One day Vitaly was going home late at night and wondering: how many people aren’t sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment.

Vitaly sees a building of n floors and 2·m windows on each floor. On each floor there are m flats numbered from 1 to m, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2·m from left to right, then the j-th flat of the i-th floor has windows 2·j?-?1 and 2·j in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren’t sleeping at that moment if at least one of the windows corresponding to this flat has lights on.

Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren’t sleeping.

Input
The first line of the input contains two integers n and m (1?≤?n,?m?≤?100) — the number of floors in the house and the number of flats on each floor respectively.

Next n lines describe the floors from top to bottom and contain 2·m characters each. If the i-th window of the given floor has lights on, then the i-th character of this line is ‘1’, otherwise it is ‘0’.

Output
Print a single integer — the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren’t sleeping.

Examples
Input
2 2
0 0 0 1
1 0 1 1
Output
3
Input
1 3
1 1 0 1 0 0
Output
2
Note
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn’t on only on the second floor in the left flat. That is, in both rooms of the flat the light is off.

In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
C語(yǔ)言版本一

#include <stdio.h> #include <stdlib.h>/* run this program using the console pauser or add your own getch, system("pause") or input loop */int main(int argc, char *argv[]) {int n,m;scanf("%d%d",&n,&m);int d[110][210];int i,j;for(i=0;i<n;i++){for(j=0;j<2*m;j++){scanf("%d",&d[i][j]);}}int count=0;for(i=0;i<n;i++){for(j=0;j<2*m;j++){if(d[i][j]==1||d[i][j+1]==1)count+=1;j++;}}printf("%d\n",count);return 0; }

C++版本一

#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <algorithm> #include <queue> #include <stack> #include <map> #include <vector> #define INF 0x3f3f3f3f #define eps 1e-8 #define MAXN (200+10) #define MAXM (50000000) #define Ri(a) scanf("%d", &a) #define Rl(a) scanf("%lld", &a) #define Rf(a) scanf("%lf", &a) #define Rs(a) scanf("%s", a) #define Pi(a) printf("%d\n", (a)) #define Pf(a) printf("%lf\n", (a)) #define Pl(a) printf("%lld\n", (a)) #define Ps(a) printf("%s\n", (a)) #define W(a) while(a--) #define CLR(a, b) memset(a, (b), sizeof(a)) #define MOD 1000000007 #define LL long long #define lson o<<1, l, mid #define rson o<<1|1, mid+1, r #define ll o<<1 #define rr o<<1|1 using namespace std; int main() {int n, m;Ri(n); Ri(m);int ans = 0;for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){int a, b; Ri(a); Ri(b);if(a || b)ans++;}}Pi(ans);return 0; }

C++版本二

#include<stdio.h> #include<iostream> #include<string.h> #include<string> #include<ctype.h> #include<math.h> #include<set> #include<map> #include<vector> #include<queue> #include<bitset> #include<algorithm> #include<time.h> using namespace std; void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);} #define MS(x,y) memset(x,y,sizeof(x)) #define MC(x,y) memcpy(x,y,sizeof(x)) #define MP(x,y) make_pair(x,y) #define ls o<<1 #define rs o<<1|1 typedef long long LL; typedef unsigned long long UL; typedef unsigned int UI; template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;} template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;} const int N=0,M=0,Z=1e9+7,ms63=1061109567; int n,m; int x,y; int main() {while(~scanf("%d%d",&n,&m)){int ans=0;for(int i=1;i<=n;++i){for(int j=1;j<=m;++j){scanf("%d%d",&x,&y);ans+=x+y>0;}}printf("%d\n",ans);}return 0; } /* 【題意】 給你n層樓（floor），每層m個(gè)房間（flat），每個(gè)房間2間窗戶。亮燈用1表示，不亮燈用0表示。 一間房間亮燈則代表這間房間有人。讓你輸出有多少間房間有人。 【類(lèi)型】 水題 【分析】 直接暴力…… */

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