{"id":17,"date":"2022-06-04T10:52:00","date_gmt":"2022-06-04T02:52:00","guid":{"rendered":"http:\/\/ubuntu.tim-wcx.ltd\/wordpress\/index.php\/2022\/06\/04\/2020-09-11-%e6%a0%91%e7%8a%b6%e6%95%b0%e7%bb%84%e6%b1%82%e6%9c%80%e5%a4%a7%e5%80%bc\/"},"modified":"2022-06-04T10:52:00","modified_gmt":"2022-06-04T02:52:00","slug":"2020-09-11-%e6%a0%91%e7%8a%b6%e6%95%b0%e7%bb%84%e6%b1%82%e6%9c%80%e5%a4%a7%e5%80%bc","status":"publish","type":"post","link":"https:\/\/wordpress.tim-wcx.ltd\/index.php\/2022\/06\/04\/2020-09-11-%e6%a0%91%e7%8a%b6%e6%95%b0%e7%bb%84%e6%b1%82%e6%9c%80%e5%a4%a7%e5%80%bc\/","title":{"rendered":"2020-09-11 \u6811\u72b6\u6570\u7ec4\u6c42\u6700\u5927\u503c"},"content":{"rendered":"

\n <\/path>\n <\/svg><\/p>\n

http:\/\/icpc.upc.edu.cn\/status.php?user_id=2019UPC135&cid=1461<\/a><\/p>\n

<\/a>\u9898\u76ee\u63cf\u8ff0<\/h5>\n

\u73b0\u5728\u8bf7\u6c42\u4f60\u7ef4\u62a4\u4e00\u4e2a\u6570\u5217\uff0c\u8981\u6c42\u63d0\u4f9b\u4ee5\u4e0b\u4e24\u79cd\u64cd\u4f5c\uff1a1\u3001\u67e5\u8be2\u64cd\u4f5c\u3002\u8bed\u6cd5\uff1aQL\u529f\u80fd\uff1a\u67e5\u8be2\u5f53\u524d\u6570\u5217\u4e2d\u672b\u5c3eL\u4e2a\u6570\u4e2d\u7684\u6700\u5927\u7684\u6570\uff0c\u5e76\u8f93\u51fa\u8fd9\u4e2a\u6570\u7684\u503c\u3002\u9650\u5236\uff1aL\u4e0d\u8d85\u8fc7\u5f53\u524d\u6570\u5217\u7684\u957f\u5ea6\u30022\u3001\u63d2\u5165\u64cd\u4f5c\u3002\u8bed\u6cd5\uff1aAn\u529f\u80fd\uff1a\u5c06n\u52a0\u4e0at\uff0c\u5176\u4e2dt\u662f\u6700\u8fd1\u4e00\u6b21\u67e5\u8be2\u64cd\u4f5c\u7684\u7b54\u6848\uff08\u5982\u679c\u8fd8\u672a\u6267\u884c\u8fc7\u67e5\u8be2\u64cd\u4f5c\uff0c\u5219t=0)\uff0c\u5e76\u5c06\u6240\u5f97\u7ed3\u679c\u5bf9\u4e00\u4e2a\u56fa\u5b9a\u7684\u5e38\u6570D\u53d6\u6a21\uff0c\u5c06\u6240\u5f97\u7b54\u6848\u63d2\u5165\u5230\u6570\u5217\u7684\u672b\u5c3e\u3002\u9650\u5236\uff1an\u662f\u975e\u8d1f\u6574\u6570\u5e76\u4e14\u5728\u957f\u6574\u8303\u56f4\u5185\u3002\u6ce8\u610f\uff1a\u521d\u59cb\u65f6\u6570\u5217\u662f\u7a7a\u7684\uff0c\u6ca1\u6709\u4e00\u4e2a\u6570\u3002<\/p>\n

<\/a>\u8f93\u5165<\/h5>\n

\u7b2c\u4e00\u884c\u4e24\u4e2a\u6574\u6570\uff0cM\u548cD\uff0c\u5176\u4e2dM\u8868\u793a\u64cd\u4f5c\u7684\u4e2a\u6570(M <= 200,000)\uff0cD\u5982\u4e0a\u6587\u4e2d\u6240\u8ff0\uff0c\u6ee1\u8db3D\u5728longint\u5185\u3002\u63a5\u4e0b\u6765M\u884c\uff0c\u67e5\u8be2\u64cd\u4f5c\u6216\u8005\u63d2\u5165\u64cd\u4f5c\u3002<\/p>\n

<\/a>\u8f93\u51fa<\/h5>\n

\u5bf9\u4e8e\u6bcf\u4e00\u4e2a\u8be2\u95ee\u64cd\u4f5c\uff0c\u8f93\u51fa\u4e00\u884c\u3002\u8be5\u884c\u53ea\u6709\u4e00\u4e2a\u6570\uff0c\u5373\u5e8f\u5217\u4e2d\u6700\u540eL\u4e2a\u6570\u7684\u6700\u5927\u6570\u3002<\/p>\n

<\/a>\u6837\u4f8b\u8f93\u5165<\/h5>\n
5 100\nA 96\nQ 1\nA 97\nQ 1\nQ 2\n<\/code><\/pre>\n
<\/a>\u6837\u4f8b\u8f93\u51fa<\/h5>\n
96\n93\n96\n<\/code><\/pre>\n
\u6811\u72b6\u6570\u7ec4\u7ef4\u62a4\u6570\u7ec4\u6c42\u6700\u503c
 \u9644\u4e00\u5f20\u56fe\u5e2e\u52a9\u7406\u89e3\u4ee3\u7801
 \"\u5728\u8fd9\u91cc\u63d2\u5165\u56fe\u7247\u63cf\u8ff0\"<\/p>\n
#pragma<\/span> GCC optimize(1)<\/span>\n#pragma<\/span> GCC optimize(2)<\/span>\n#pragma<\/span> GCC optimize(3,\"Ofast\",\"inline\")<\/span>\n#include<\/span><\/span><\/span>\n#define<\/span> inf 0x3f3f3f<\/span>\nusing<\/span> namespace<\/span> std;<\/span>\ntypedef<\/span> long<\/span> long<\/span> ll;<\/span>\nll c[<\/span>400500<\/span>]<\/span>=<\/span>{<\/span>0<\/span>}<\/span>;<\/span>\nll h[<\/span>400500<\/span>]<\/span>=<\/span>{<\/span>0<\/span>}<\/span>;<\/span>\nll lowbit<\/span>(<\/span>ll x)<\/span>\n{<\/span>\n    return<\/span> x&<\/span>(<\/span>-<\/span>x)<\/span>;<\/span>\n}<\/span>\nll getmax<\/span>(<\/span>ll l,<\/span>ll r)<\/span>\n{<\/span>\n    ll max1=<\/span>h[<\/span>r]<\/span>;<\/span>\n    while<\/span>(<\/span>l<=<\/span>r)<\/span>\n    {<\/span>\n        max1=<\/span>max<\/span>(<\/span>max1,<\/span>h[<\/span>r]<\/span>)<\/span>;<\/span>\n        for<\/span>(<\/span>--<\/span>r;<\/span>r-<\/span>l>=<\/span>lowbit<\/span>(<\/span>r)<\/span>;<\/span>r-<\/span>=<\/span>lowbit<\/span>(<\/span>r)<\/span>)<\/span>\n            max1=<\/span>max<\/span>(<\/span>max1,<\/span>c[<\/span>r]<\/span>)<\/span>;<\/span>\n    }<\/span>\n    return<\/span> max1;<\/span>\n}<\/span>\nchar<\/span> s[<\/span>100<\/span>]<\/span>;<\/span>\nint<\/span> main<\/span>(<\/span>)<\/span>\n{<\/span>\n    ll n,<\/span>d,<\/span>a,<\/span>t=<\/span>0<\/span>,<\/span>cnt=<\/span>0<\/span>;<\/span>\n    scanf<\/span>(<\/span>\"%lld%lld\"<\/span>,<\/span>&<\/span>n,<\/span>&<\/span>d)<\/span>;<\/span>\n    while<\/span>(<\/span>n--<\/span>)<\/span>\n    {<\/span>\n        scanf<\/span>(<\/span>\"%s%lld\"<\/span>,<\/span>s,<\/span>&<\/span>a)<\/span>;<\/span>\n        if<\/span>(<\/span>s[<\/span>0<\/span>]<\/span>==<\/span>'A'<\/span>)<\/span>\n        {<\/span>\n            h[<\/span>++<\/span>cnt]<\/span>=<\/span>(<\/span>a+<\/span>t)<\/span>%<\/span>d;<\/span>\n            c[<\/span>cnt]<\/span>=<\/span>max<\/span>(<\/span>getmax<\/span>(<\/span>cnt-<\/span>lowbit<\/span>(<\/span>cnt)<\/span>+<\/span>1<\/span>,<\/span>cnt-<\/span>1<\/span>)<\/span>,<\/span>h[<\/span>cnt]<\/span>)<\/span>;<\/span>\n        }<\/span>\n        else<\/span>\n        {<\/span>\n            t=<\/span>getmax<\/span>(<\/span>cnt-<\/span>a+<\/span>1<\/span>,<\/span>cnt)<\/span>;<\/span>\n            printf<\/span>(<\/span>\"%dn\"<\/span>,<\/span>t)<\/span>;<\/span>\n        }<\/span>\n    }<\/span>\n}<\/span>\n<\/code><\/pre>\n
<\/a>\u5355\u70b9\u4fee\u6539+\u533a\u95f4\u67e5\u8be2\u6c42\u548c<\/h5>\n
<\/a>\u9898\u76ee<\/h6>\n
\u7ed9\u5b9a\u4e00\u4e2a\u6570\u5b57n\uff0c\u8868\u793a\u6570\u7ec4\u5927\u5c0f\uff0c\u540e\u6709n\u4e2a\u6570\u5b57\uff0c\u8868\u793a\u6570\u7ec4\u4e2d\u6bcf\u4e2a\u6570\u5b57\u7684\u521d\u59cb\u503c\uff0c\u7136\u540e\u8fdb\u884cm\u6b21\u64cd\u4f5c\uff0c\u2018Q\u2019\u4ee3\u8868\u67e5\u8be2\uff0c\u67e5\u8be2\u67d0\u4e00\u533a\u95f4\u6240\u6709\u5143\u7d20\u7684\u548c\uff0c\u2018C\u2019\u4ee3\u8868\u4fee\u6539\uff0c\u5728\u67d0\u4e00\u5143\u7d20\u4e0a\u52a0\u4e0a\u503c\u3002<\/p>\n
<\/a>\u8f93\u5165<\/h6>\n
10 7\n1 2 3 4 5 6 7 8 9 10\nQ 4 4\nQ 1 10\nQ 2 4\nC 3 6\nQ 2 4\nC 2 4\nQ 2 4\n<\/code><\/pre>\n
<\/a>\u8f93\u51fa<\/h6>\n
4\n55\n9\n15\n18\n<\/code><\/pre>\n
\u6811\u72b6\u6570\u7ec4\u5355\u70b9\u4fee\u6539+\u533a\u95f4\u6c42\u548c<\/p>\n
#include<\/span> <\/span><\/span>\n#include<\/span> <\/span><\/span>\nusing<\/span> namespace<\/span> std;<\/span>\ntypedef<\/span> long<\/span> long<\/span> ll;<\/span>\nll n,<\/span>m;<\/span>\nll h[<\/span>100500<\/span>]<\/span>=<\/span>{<\/span>0<\/span>}<\/span>;<\/span>\nll c[<\/span>100500<\/span>]<\/span>=<\/span>{<\/span>0<\/span>}<\/span>;<\/span>\nll lowbit<\/span>(<\/span>ll x)<\/span>\n{<\/span>\n    return<\/span> x&<\/span>(<\/span>-<\/span>x)<\/span>;<\/span>\n}<\/span>\nll add<\/span>(<\/span>ll x,<\/span>ll a)<\/span>\n{<\/span>\n    while<\/span>(<\/span>x<=<\/span>n)<\/span>\n    {<\/span>\n        c[<\/span>x]<\/span>+<\/span>=<\/span>a;<\/span>\n        x+<\/span>=<\/span>lowbit<\/span>(<\/span>x)<\/span>;<\/span>\n    }<\/span>\n}<\/span>\nll getsum<\/span>(<\/span>ll l)<\/span>\n{<\/span>\n    ll sum=<\/span>0<\/span>;<\/span>\n    while<\/span>(<\/span>l><\/span>0<\/span>)<\/span>\n    {<\/span>\n        sum+<\/span>=<\/span>c[<\/span>l]<\/span>;<\/span>\n        l-<\/span>=<\/span>lowbit<\/span>(<\/span>l)<\/span>;<\/span>\n    }<\/span>\n    return<\/span> sum;<\/span>\n}<\/span>\nint<\/span> main<\/span>(<\/span>)<\/span>\n{<\/span>\n    scanf<\/span>(<\/span>\"%lld%lld\"<\/span>,<\/span>&<\/span>n,<\/span>&<\/span>m)<\/span>;<\/span>\n    for<\/span>(<\/span>int<\/span> i=<\/span>1<\/span>;<\/span>i<=<\/span>n;<\/span>i++<\/span>)<\/span>\n    {<\/span>\n        scanf<\/span>(<\/span>\"%lld\"<\/span>,<\/span>&<\/span>h[<\/span>i]<\/span>)<\/span>;<\/span>\n        add<\/span>(<\/span>h[<\/span>i]<\/span>,<\/span>i)<\/span>;<\/span>\n    }<\/span>\n    char<\/span> a[<\/span>100<\/span>]<\/span>;<\/span>\n    ll b,<\/span>c;<\/span>\n    while<\/span>(<\/span>m--<\/span>)<\/span>\n    {<\/span>\n        scanf<\/span>(<\/span>\"%s%lld%lld\"<\/span>,<\/span>a,<\/span>&<\/span>b,<\/span>&<\/span>c)<\/span>;<\/span>\n        if<\/span>(<\/span>a[<\/span>0<\/span>]<\/span>==<\/span>'Q'<\/span>)<\/span>\n        {<\/span>\n            printf<\/span>(<\/span>\"%lldn\"<\/span>,<\/span>getsum<\/span>(<\/span>c)<\/span>-<\/span>getsum<\/span>(<\/span>b-<\/span>1<\/span>)<\/span>)<\/span>;<\/span>\n        }<\/span>\n        else<\/span>\n        {<\/span>\n            add<\/span>(<\/span>b,<\/span>c)<\/span>;<\/span>\n        }<\/span>\n    }<\/span>\n    return<\/span> 0<\/span>;<\/span>\n}<\/span>\n<\/code><\/pre>\n
<\/a>\u533a\u95f4\u4fee\u6539+\u533a\u95f4\u67e5\u8be2<\/h5>\n
http:\/\/icpc.upc.edu.cn\/problem.php?cid=1461&pid=13<\/a><\/p>\n
<\/a>\u9898\u76ee\u63cf\u8ff0<\/h6>\n
You have N integers, A1, A2, \u2026 , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.<\/p>\n
<\/a>\u8f93\u5165<\/h6>\n
The first line contains two numbers N and Q. 1 \u2264 N,Q \u2264 100000.
 The second line contains N numbers, the initial values of A1, A2, \u2026 , AN. -1000000000 \u2264 Ai \u2264 1000000000.
 Each of the next Q lines represents an operation.
 \u201cC a b c\u201d means adding c to each of Aa, Aa+1, \u2026 , Ab. -10000 \u2264 c \u2264 10000.
 \u201cQ a b\u201d means querying the sum of Aa, Aa+1, \u2026 , Ab.<\/p>\n
<\/a>\u8f93\u51fa<\/h6>\n
You need to answer all Q commands in order. One answer in a line.<\/p>\n
<\/a>\u6837\u4f8b\u8f93\u5165<\/h6>\n
10 5\n1 2 3 4 5 6 7 8 9 10\nQ 4 4\nQ 1 10\nQ 2 4\nC 3 6 3\nQ 2 4\n<\/code><\/pre>\n
<\/a>\u6837\u4f8b\u8f93\u51fa<\/h6>\n
4\n55\n9\n15\n<\/code><\/pre>\n
\u63d0\u793a
 The sums may exceed the range of 32-bit integers.<\/p>\n
#include<\/span><\/span><\/span>\n#define<\/span> inf 0x3f3f3f<\/span>\nusing<\/span> namespace<\/span> std;<\/span>\ntypedef<\/span> long<\/span> long<\/span> ll;<\/span>\nll num[<\/span>400500<\/span>]<\/span>=<\/span>{<\/span>0<\/span>}<\/span>;<\/span>\nll sum[<\/span>400500<\/span>]<\/span>=<\/span>{<\/span>0<\/span>}<\/span>;<\/span>\nll lazy[<\/span>400500<\/span>]<\/span>=<\/span>{<\/span>0<\/span>}<\/span>;<\/span>\nll n,<\/span>m,<\/span>b,<\/span>c,<\/span>d;<\/span>\nll pushup<\/span>(<\/span>ll t)<\/span>\n{<\/span>\n    sum[<\/span>t]<\/span>=<\/span>sum[<\/span>2<\/span>*<\/span>t]<\/span>+<\/span>sum[<\/span>2<\/span>*<\/span>t+<\/span>1<\/span>]<\/span>;<\/span>\n}<\/span>\nll pushlazy<\/span>(<\/span>ll t,<\/span>ll lz,<\/span>ll len)<\/span>\n{<\/span>\n    sum[<\/span>t]<\/span>+<\/span>=<\/span>lz*<\/span>len;<\/span>\n    lazy[<\/span>t]<\/span>+<\/span>=<\/span>lz;<\/span>\n}<\/span>\nll pushdown<\/span>(<\/span>ll l,<\/span>ll r,<\/span>ll t)<\/span>\n{<\/span>\n    if<\/span>(<\/span>lazy[<\/span>t]<\/span>!=<\/span>0<\/span>)<\/span>\n    {<\/span>\n        ll mid=<\/span>(<\/span>l+<\/span>r)<\/span>\/<\/span>2<\/span>;<\/span>\n        pushlazy<\/span>(<\/span>2<\/span>*<\/span>t,<\/span>lazy[<\/span>t]<\/span>,<\/span>mid-<\/span>l+<\/span>1<\/span>)<\/span>;<\/span>\n        pushlazy<\/span>(<\/span>2<\/span>*<\/span>t+<\/span>1<\/span>,<\/span>lazy[<\/span>t]<\/span>,<\/span>r-<\/span>mid)<\/span>;<\/span>\n        lazy[<\/span>t]<\/span>=<\/span>0<\/span>;<\/span>\n    }<\/span>\n}<\/span>\nll build<\/span>(<\/span>ll l,<\/span>ll r,<\/span>ll t)<\/span>\n{<\/span>\n    lazy[<\/span>t]<\/span>=<\/span>0<\/span>;<\/span>\n    if<\/span>(<\/span>l==<\/span>r)<\/span>\n    {<\/span>\n        sum[<\/span>t]<\/span>=<\/span>num[<\/span>l]<\/span>;<\/span>\n        return<\/span> 0<\/span>;<\/span>\n    }<\/span>\n    ll mid=<\/span>(<\/span>l+<\/span>r)<\/span>\/<\/span>2<\/span>;<\/span>\n    build<\/span>(<\/span>l,<\/span>mid,<\/span>t*<\/span>2<\/span>)<\/span>;<\/span>\n    build<\/span>(<\/span>mid+<\/span>1<\/span>,<\/span>r,<\/span>2<\/span>*<\/span>t+<\/span>1<\/span>)<\/span>;<\/span>\n    pushup<\/span>(<\/span>t)<\/span>;<\/span>\n}<\/span>\nll add<\/span>(<\/span>ll x,<\/span>ll l,<\/span>ll r,<\/span>ll L,<\/span>ll R,<\/span>ll t)<\/span>\n{<\/span>\n    if<\/span>(<\/span>l>=<\/span>L&&<\/span>r<=<\/span>R)<\/span>\n    {<\/span>\n        lazy[<\/span>t]<\/span>+<\/span>=<\/span>x;<\/span>\n        sum[<\/span>t]<\/span>+<\/span>=<\/span>(<\/span>r-<\/span>l+<\/span>1<\/span>)<\/span>*<\/span>x;<\/span>\n        return<\/span> 0<\/span>;<\/span>\n    }<\/span>\n    pushdown<\/span>(<\/span>l,<\/span>r,<\/span>t)<\/span>;<\/span>\n    ll mid=<\/span>(<\/span>l+<\/span>r)<\/span>\/<\/span>2<\/span>;<\/span>\n    if<\/span>(<\/span>L<=<\/span>mid)<\/span>\n        add<\/span>(<\/span>x,<\/span>l,<\/span>mid,<\/span>L,<\/span>R,<\/span>2<\/span>*<\/span>t)<\/span>;<\/span>\n    if<\/span>(<\/span>R><\/span>mid)<\/span>\n        add<\/span>(<\/span>x,<\/span>mid+<\/span>1<\/span>,<\/span>r,<\/span>L,<\/span>R,<\/span>2<\/span>*<\/span>t+<\/span>1<\/span>)<\/span>;<\/span>\n    pushup<\/span>(<\/span>t)<\/span>;<\/span>\n}<\/span>\nll querysum<\/span>(<\/span>ll l,<\/span>ll r,<\/span>ll L,<\/span>ll R,<\/span>ll t)<\/span>\n{<\/span>\n    if<\/span>(<\/span>l>=<\/span>L&&<\/span>r<=<\/span>R)<\/span>\n        return<\/span> sum[<\/span>t]<\/span>;<\/span>\n    pushdown<\/span>(<\/span>l,<\/span>r,<\/span>t)<\/span>;<\/span>\n    ll mid=<\/span>(<\/span>l+<\/span>r)<\/span>\/<\/span>2<\/span>;<\/span>\n    ll ans=<\/span>0<\/span>;<\/span>\n    if<\/span>(<\/span>L<=<\/span>mid)<\/span>\n        ans+<\/span>=<\/span>querysum<\/span>(<\/span>l,<\/span>mid,<\/span>L,<\/span>R,<\/span>2<\/span>*<\/span>t)<\/span>;<\/span>\n    if<\/span>(<\/span>R><\/span>mid)<\/span>\n        ans+<\/span>=<\/span>querysum<\/span>(<\/span>mid+<\/span>1<\/span>,<\/span>r,<\/span>L,<\/span>R,<\/span>2<\/span>*<\/span>t+<\/span>1<\/span>)<\/span>;<\/span>\n    return<\/span> ans;<\/span>\n}<\/span>\nint<\/span> main<\/span>(<\/span>)<\/span>\n{<\/span>\n    scanf<\/span>(<\/span>\"%lld%lld\"<\/span>,<\/span>&<\/span>n,<\/span>&<\/span>m)<\/span>;<\/span>\n    for<\/span>(<\/span>ll i=<\/span>1<\/span>;<\/span>i<=<\/span>n;<\/span>i++<\/span>)<\/span>\n        scanf<\/span>(<\/span>\"%lld\"<\/span>,<\/span>&<\/span>num[<\/span>i]<\/span>)<\/span>;<\/span>\n    build<\/span>(<\/span>1<\/span>,<\/span>n,<\/span>1<\/span>)<\/span>;<\/span>\n    char<\/span> a[<\/span>100<\/span>]<\/span>=<\/span>{<\/span>0<\/span>}<\/span>;<\/span>\n    for<\/span>(<\/span>ll i=<\/span>1<\/span>;<\/span>i<=<\/span>m;<\/span>i++<\/span>)<\/span>\n    {<\/span>\n        scanf<\/span>(<\/span>\"%s\"<\/span>,<\/span>a)<\/span>;<\/span>\n        if<\/span>(<\/span>a[<\/span>0<\/span>]<\/span>==<\/span>'C'<\/span>)<\/span>\n        {<\/span>\n            scanf<\/span>(<\/span>\"%lld%lld%lld\"<\/span>,<\/span>&<\/span>b,<\/span>&<\/span>c,<\/span>&<\/span>d)<\/span>;<\/span>\n            add<\/span>(<\/span>d,<\/span>1<\/span>,<\/span>n,<\/span>b,<\/span>c,<\/span>1<\/span>)<\/span>;<\/span>\n        }<\/span>\n        else<\/span>\n        {<\/span>\n            scanf<\/span>(<\/span>\"%lld%lld\"<\/span>,<\/span>&<\/span>b,<\/span>&<\/span>c)<\/span>;<\/span>\n            printf<\/span>(<\/span>\"%lldn\"<\/span>,<\/span>querysum<\/span>(<\/span>1<\/span>,<\/span>n,<\/span>b,<\/span>c,<\/span>1<\/span>)<\/span>)<\/span>;<\/span>\n        }<\/span>\n    }<\/span>\n}<\/span>\n<\/code><\/pre>\n","protected":false},"excerpt":{"rendered":"
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